Formal geometry proofs

1. Proofs of triangle and polygon theorems

ID is: 1857 Seed is: 1074

Interior and exterior angles of a triangle

In the triangle below, the interior angles are given as $\angle M =$ $\text{90}°$ , $\angle N = x$ , and $\angle P = \text{45}°$ . Sides $\overline{MN}$ and $\overline{MP}$ are equal in length, as indicated. An exterior angle $y$ is formed by extending one side of the triangle toward point $Q$ .

Calculate the size of the interior angle, $x$ .
Calculate the size of the exterior angle, $y$ .

INSTRUCTION: Your answers should be exact (do not round off).

Answer:

$x =$ $°$
$y =$ $°$

numeric

2 attempts remaining

HINT: <no title>
[−1 point ⇒ 1 / 2 points left]

Start with the fact that the interior angles of a triangle sum to $180°$ .

STEP: Use the sum of angles in a triangle to find

x

$x$
[−1 point ⇒ 0 / 2 points left]

We need to find two different unknown angles in the diagram. We can find them using facts we know about triangles.

One of the unknown angles, $x$ , is inside of the triangle. So we can use the fact that the interior angles of a triangle have a sum of $180°$ . Write an equation for this and then solve for $x$ .

\begin{aligned} Interior angles of a triangle & = 180 ° \\ 90 ° + x + 45 ° & = 180 ° \end{aligned}

$\begin{aligned} \text{Interior angles of a triangle} & = 180° \\ \text{90}° + x + \text{45}° & = 180° \end{aligned}$

\begin{aligned} x & = 180 ° - 90 ° - 45 ° \\ x & = 45 ° \end{aligned}

$\begin{aligned} x & = 180° - \text{90}° - \text{45}° \\ x & = \text{45} {°} \end{aligned}$

NOTE: You can also solve for

x

$x$ using the fact that this triangle is isosceles (it has two equal sides). Isosceles triangles have two congruent (equal) angles. In this case the equal angles are

∠ N

$\angle N$ and

∠ P

$\angle P$ . And that means that

x

$x$ is equal to

45 °

$\text{45}°$ .

Angle $x$ has a measure of $\text{45}°$ .

STEP: Use the exterior angle theorem to find

y

$y$
[−1 point ⇒ 0 / 2 points left]

The other angle we need to find, $y$ , is outside of the triangle. It is an exterior angle because it is made by the extension of one of the sides of the triangle. To solve for $y$ use the theorem for the exterior angles of a triangle.

The exterior angle theorem for triangles tells us that an exterior angle is equal to the sum of the two interior angles opposite the exterior angle. In this triangle, the exterior angle $Q\hat{M}P$ is equal to the sum of the opposite interior angles, $\angle N$ and $\angle P$ . The figure below shows this with shaded angles. If we add the orange and blue angles inside the triangle together, we will get an angle which is exactly the same size as the exterior angle.

We can write an equation based on the exterior angle theorem to find the value of $y$ :

Exterior angle of a triangle = The sum of the opposite interior angles

\begin{aligned} y & = ∠ N + ∠ P \\ y & = 45 ° + 45 ° \\ y & = 90 ° \end{aligned}

$\begin{aligned} y & = \angle N + \angle P \\ y & = \text{45}° + \text{45}° \\ y & = \text{90}{°} \end{aligned}$

NOTE: You can also find

y

$y$ using supplementary angles. The two angles at

M

$M$ must have a sum of

180 °

$180°$ because they make a straight line. This approach will lead to the same answer,

y = 90 °

$y = \text{90}°$ .

The correct answers are:

$x =$ $\text{45}$ ${°}$
$y =$ $\text{90}$ ${°}$

Submit your answer as: and

ID is: 1857 Seed is: 9480

Interior and exterior angles of a triangle

In the triangle below, the interior angles are given as $\angle M =$ $\text{83.62}°$ , $\angle N = \text{48.19}°$ , and $\angle P = x$ . Sides $\overline{MN}$ and $\overline{MP}$ are equal in length, as indicated. An exterior angle $y$ is formed by extending one side of the triangle toward point $Q$ .

Calculate the size of the interior angle, $x$ .
Calculate the size of the exterior angle, $y$ .

INSTRUCTION: Your answers should be exact (do not round off).

Answer:

$x =$ $°$
$y =$ $°$

numeric

2 attempts remaining

HINT: <no title>
[−1 point ⇒ 1 / 2 points left]

Start with the fact that the interior angles of a triangle sum to $180°$ .

STEP: Use the sum of angles in a triangle to find

x

$x$
[−1 point ⇒ 0 / 2 points left]

We need to find two different unknown angles in the diagram. We can find them using facts we know about triangles.

One of the unknown angles, $x$ , is inside of the triangle. So we can use the fact that the interior angles of a triangle have a sum of $180°$ . Write an equation for this and then solve for $x$ .

\begin{aligned} Interior angles of a triangle & = 180 ° \\ 83.62 ° + 48.19 ° + x & = 180 ° \end{aligned}

$\begin{aligned} \text{Interior angles of a triangle} & = 180° \\ \text{83.62}° + \text{48.19}° + x & = 180° \end{aligned}$

\begin{aligned} x & = 180 ° - 83.62 ° - 48.19 ° \\ x & = 48.19 ° \end{aligned}

$\begin{aligned} x & = 180° - \text{83.62}° - \text{48.19}° \\ x & = \text{48.19} {°} \end{aligned}$

NOTE: You can also solve for

x

$x$ using the fact that this triangle is isosceles (it has two equal sides). Isosceles triangles have two congruent (equal) angles. In this case the equal angles are

∠ N

$\angle N$ and

∠ P

$\angle P$ . And that means that

x

$x$ is equal to

48.19 °

$\text{48.19}°$ .

Angle $x$ has a measure of $\text{48.19}°$ .

STEP: Use the exterior angle theorem to find

y

$y$
[−1 point ⇒ 0 / 2 points left]

We can write an equation based on the exterior angle theorem to find the value of $y$ :

Exterior angle of a triangle = The sum of the opposite interior angles

\begin{aligned} y & = ∠ N + ∠ P \\ y & = 48.19 ° + 48.19 ° \\ y & = 96.38 ° \end{aligned}

$\begin{aligned} y & = \angle N + \angle P \\ y & = \text{48.19}° + \text{48.19}° \\ y & = \text{96.38}{°} \end{aligned}$

NOTE: You can also find

y

$y$ using supplementary angles. The two angles at

M

$M$ must have a sum of

180 °

$180°$ because they make a straight line. This approach will lead to the same answer,

y = 96.38 °

$y = \text{96.38}°$ .

The correct answers are:

$x =$ $\text{48.19}$ ${°}$
$y =$ $\text{96.38}$ ${°}$

Submit your answer as: and

ID is: 1857 Seed is: 6710

Interior and exterior angles of a triangle

In the triangle below, the interior angles are given as $\angle M =$ $\text{90}°$ , $\angle N = \text{45}°$ , and $\angle P = x$ . Sides $\overline{MN}$ and $\overline{MP}$ are equal in length, as indicated. An exterior angle $y$ is formed by extending one side of the triangle toward point $Q$ .

Calculate the size of the interior angle, $x$ .
Calculate the size of the exterior angle, $y$ .

INSTRUCTION: Your answers should be exact (do not round off).

Answer:

$x =$ $°$
$y =$ $°$

numeric

2 attempts remaining

HINT: <no title>
[−1 point ⇒ 1 / 2 points left]

Start with the fact that the interior angles of a triangle sum to $180°$ .

STEP: Use the sum of angles in a triangle to find

x

$x$
[−1 point ⇒ 0 / 2 points left]

We need to find two different unknown angles in the diagram. We can find them using facts we know about triangles.

One of the unknown angles, $x$ , is inside of the triangle. So we can use the fact that the interior angles of a triangle have a sum of $180°$ . Write an equation for this and then solve for $x$ .

\begin{aligned} Interior angles of a triangle & = 180 ° \\ 90 ° + 45 ° + x & = 180 ° \end{aligned}

$\begin{aligned} \text{Interior angles of a triangle} & = 180° \\ \text{90}° + \text{45}° + x & = 180° \end{aligned}$

\begin{aligned} x & = 180 ° - 90 ° - 45 ° \\ x & = 45 ° \end{aligned}

$\begin{aligned} x & = 180° - \text{90}° - \text{45}° \\ x & = \text{45} {°} \end{aligned}$

NOTE: You can also solve for

x

$x$ using the fact that this triangle is isosceles (it has two equal sides). Isosceles triangles have two congruent (equal) angles. In this case the equal angles are

∠ N

$\angle N$ and

∠ P

$\angle P$ . And that means that

x

$x$ is equal to

45 °

$\text{45}°$ .

Angle $x$ has a measure of $\text{45}°$ .

STEP: Use the exterior angle theorem to find

y

$y$
[−1 point ⇒ 0 / 2 points left]

We can write an equation based on the exterior angle theorem to find the value of $y$ :

Exterior angle of a triangle = The sum of the opposite interior angles

\begin{aligned} y & = ∠ N + ∠ P \\ y & = 45 ° + 45 ° \\ y & = 90 ° \end{aligned}

$\begin{aligned} y & = \angle N + \angle P \\ y & = \text{45}° + \text{45}° \\ y & = \text{90}{°} \end{aligned}$

NOTE: You can also find

y

$y$ using supplementary angles. The two angles at

M

$M$ must have a sum of

180 °

$180°$ because they make a straight line. This approach will lead to the same answer,

y = 90 °

$y = \text{90}°$ .

The correct answers are:

$x =$ $\text{45}$ ${°}$
$y =$ $\text{90}$ ${°}$

Submit your answer as: and

ID is: 1907 Seed is: 9473

Interior angles of polygons

Look at the following regular polygon and answer the questions that follow:

Answer: NOTE: Round your answer to two decimal places where necessary.

1. What is the sum of the interior angles of this polygon? ${°}$
2. What is the size of each interior angle of this polygon? ${°}$

numeric

2 attempts remaining

STEP: <no title>
[−4 points ⇒ 0 / 4 points left]

The diagram below represents a polygon with 8 sides. This polygon is called a octagon.

The sum of the interior angles of this octagon = 1080 ${°}$ .

\begin{aligned} Let n be the number of sides \\ ∴ n = 8 \\ Sum of Int angles of a polygon & = 180 (n - 2) \\ = 180 \times ((8) - 2) \\ = 180 \times (6) \\ Sum of Int angles of a polygon & = 1080 ° \end{aligned}

$\begin{aligned} \text{Let n be the number of sides} \\ \therefore n = 8 \\ \text{Sum of Int angles of a polygon}& = 180(n-2) \\ & = 180 \times ((8) - 2) \\ & = 180 \times (6) \\ \text{Sum of Int angles of a polygon}& = 1080 {°} \end{aligned}$

\begin{aligned} Each individual angle & = 1080 \div 8 \\ = 135 ° \end{aligned}

$\begin{aligned} \text{Each individual angle} & = 1080 \div 8 \\ & = \text{135} {°} \end{aligned}$

Submit your answer as: and

ID is: 1907 Seed is: 5984

Interior angles of polygons

Look at the following regular polygon and answer the questions that follow:

Answer: NOTE: Round your answer to two decimal places where necessary.

1. What is the sum of the interior angles of this polygon? ${°}$
2. What is the size of each interior angle of this polygon? ${°}$

numeric

2 attempts remaining

STEP: <no title>
[−4 points ⇒ 0 / 4 points left]

The diagram below represents a polygon with 6 sides. This polygon is called a hexagon.

The sum of the interior angles of this hexagon = 720 ${°}$ .

\begin{aligned} Let n be the number of sides \\ ∴ n = 6 \\ Sum of Int angles of a polygon & = 180 (n - 2) \\ = 180 \times ((6) - 2) \\ = 180 \times (4) \\ Sum of Int angles of a polygon & = 720 ° \end{aligned}

$\begin{aligned} \text{Let n be the number of sides} \\ \therefore n = 6 \\ \text{Sum of Int angles of a polygon}& = 180(n-2) \\ & = 180 \times ((6) - 2) \\ & = 180 \times (4) \\ \text{Sum of Int angles of a polygon}& = 720 {°} \end{aligned}$

\begin{aligned} Each individual angle & = 720 \div 6 \\ = 120 ° \end{aligned}

$\begin{aligned} \text{Each individual angle} & = 720 \div 6 \\ & = \text{120} {°} \end{aligned}$

Submit your answer as: and

ID is: 1907 Seed is: 9540

Interior angles of polygons

Look at the following regular polygon and answer the questions that follow:

Answer: NOTE: Round your answer to two decimal places where necessary.

1. What is the sum of the interior angles of this polygon? ${°}$
2. What is the size of each interior angle of this polygon? ${°}$

numeric

2 attempts remaining

STEP: <no title>
[−4 points ⇒ 0 / 4 points left]

The diagram below represents a polygon with 7 sides. This polygon is called a heptagon.

The sum of the interior angles of this heptagon = 900 ${°}$ .

\begin{aligned} Let n be the number of sides \\ ∴ n = 7 \\ Sum of Int angles of a polygon & = 180 (n - 2) \\ = 180 \times ((7) - 2) \\ = 180 \times (5) \\ Sum of Int angles of a polygon & = 900 ° \end{aligned}

$\begin{aligned} \text{Let n be the number of sides} \\ \therefore n = 7 \\ \text{Sum of Int angles of a polygon}& = 180(n-2) \\ & = 180 \times ((7) - 2) \\ & = 180 \times (5) \\ \text{Sum of Int angles of a polygon}& = 900 {°} \end{aligned}$

\begin{aligned} Each individual angle & = 900 \div 7 \\ = 128.57 ° \end{aligned}

$\begin{aligned} \text{Each individual angle} & = 900 \div 7 \\ & = \text{128.57} {°} \end{aligned}$

Submit your answer as: and

ID is: 1905 Seed is: 8801

Exterior angles of polygons

You are given the following regular polygon with an extension line:

What is the size of the exterior angle $x$ for this polygon?

Answer:

$x =$ ${°}$ .

numeric

2 attempts remaining

STEP: Calculate the size of the exterior angle
[−2 points ⇒ 0 / 2 points left]

The polygon shown in the question has $\text{5}$ sides. This is called a pentagon.

We are told that this is a regular polygon. That means that all of the sides have the same length, and all of the angles have the same size.

So we can use the following formula to calculate the size of the exterior angle:

\begin{aligned} Ext. angle = \frac{360 °}{n} \end{aligned}

$\begin{aligned} \text{Ext. angle} = \dfrac{360°}{n} \end{aligned}$

where $n$ is the number of sides. So

\begin{aligned} x & = \frac{360 °}{5} \\ = 72 ° \end{aligned}

$\begin{aligned} x &= \dfrac{360°}{\text{5}} \\ & = \text{72}° \end{aligned}$

Submit your answer as:

ID is: 1905 Seed is: 29

Exterior angles of polygons

You are given the following regular polygon with an extension line:

What is the size of the exterior angle $x$ for this polygon?

Answer:

$x =$ ${°}$ .

numeric

2 attempts remaining

STEP: Calculate the size of the exterior angle
[−2 points ⇒ 0 / 2 points left]

The polygon shown in the question has $\text{8}$ sides. This is called a ocatagon.

We are told that this is a regular polygon. That means that all of the sides have the same length, and all of the angles have the same size.

So we can use the following formula to calculate the size of the exterior angle:

\begin{aligned} Ext. angle = \frac{360 °}{n} \end{aligned}

$\begin{aligned} \text{Ext. angle} = \dfrac{360°}{n} \end{aligned}$

where $n$ is the number of sides. So

\begin{aligned} x & = \frac{360 °}{8} \\ = 45 ° \end{aligned}

$\begin{aligned} x &= \dfrac{360°}{\text{8}} \\ & = \text{45}° \end{aligned}$

Submit your answer as:

ID is: 1905 Seed is: 7391

Exterior angles of polygons

You are given the following regular polygon with an extension line:

What is the size of the exterior angle $x$ for this polygon?

Answer:

$x =$ ${°}$ .

numeric

2 attempts remaining

STEP: Calculate the size of the exterior angle
[−2 points ⇒ 0 / 2 points left]

The polygon shown in the question has $\text{8}$ sides. This is called a ocatagon.

We are told that this is a regular polygon. That means that all of the sides have the same length, and all of the angles have the same size.

So we can use the following formula to calculate the size of the exterior angle:

\begin{aligned} Ext. angle = \frac{360 °}{n} \end{aligned}

$\begin{aligned} \text{Ext. angle} = \dfrac{360°}{n} \end{aligned}$

where $n$ is the number of sides. So

\begin{aligned} x & = \frac{360 °}{8} \\ = 45 ° \end{aligned}

$\begin{aligned} x &= \dfrac{360°}{\text{8}} \\ & = \text{45}° \end{aligned}$

Submit your answer as:

ID is: 1908 Seed is: 414

Problems with polygons

You are given the following regular polygon with a extension line:

What is the size of exterior angle $x$ for the given polygon?

Answer:

$x =$ °

one-of

type(numeric.abserror(0.01))

2 attempts remaining

STEP: <no title>
[−2 points ⇒ 0 / 2 points left]

A plane 2D shape with straight lines that enclose, (one side follows the other side until it forms a closed shape) is a polygon. When all the sides and angles are the same size, it's called a regular polygon. The diagram below represents a polygon with $\text{8}$ sides. This polygon is called a octagon.

Let $n$ be the number of sides. Then $n = \text{8}$ .

\begin{aligned} x & = Size of Ext angle of polygon \\ = \frac{360 °}{n} \\ = \frac{360 °}{8} \\ = 45 ° \end{aligned}

$\begin{aligned} x & = \text{Size of Ext angle of polygon} \\ & = \frac{\text{360}\text{°}}{n} \\ & = \frac{\text{360}\text{°}}{8} \\ & = \text{45}\text{°} \end{aligned}$

We could also calculate the interior angle and then use supplementary rule to find $x$ .

The sum of the interior angles of a polygon is given by

180 ° (n - 2)

$\text{180}\text{°}(n-2)$ .

Therefore the sum of the angles in this polygon is:

\begin{aligned} 180 ° ((8) - 2) & = 180 ° (6) \\ = 1,080 ° \end{aligned}

$\begin{aligned} \text{180}\text{°} ((8) - 2) & = \text{180}\text{°} (6) \\ & = \text{1,080}\text{°} \end{aligned}$

We can use this to work out what each individual interior angle is.

\begin{aligned} interior angle & = 1,080 ° \div 8 \\ = 135 ° \end{aligned}

$\begin{aligned} \text{interior angle} & = \text{1,080}\text{°} \div 8 \\ & = \text{135}\text{°} \end{aligned}$

Finally, we know that the sum of angles on a straight line is $\text{180}\text{°}$ . So

\begin{aligned} exterior angle & = 180 ° - interior angle \\ = 180 ° - 135 ° \\ = 45 ° \end{aligned}

$\begin{aligned} \text{exterior angle} & = \text{180}\text{°} - \text{interior angle} \\ & = \text{180}\text{°} - \text{135}\text{°} \\ & = \text{45}\text{°} \end{aligned}$

So $x = \text{45}\text{°}$

Submit your answer as:

ID is: 1908 Seed is: 4190

Problems with polygons

You are given the following regular polygon with a extension line:

What is the size of exterior angle $x$ for the given polygon?

Answer:

$x =$ °

one-of

type(numeric.abserror(0.01))

2 attempts remaining

STEP: <no title>
[−2 points ⇒ 0 / 2 points left]

Let $n$ be the number of sides. Then $n = \text{10}$ .

\begin{aligned} x & = Size of Ext angle of polygon \\ = \frac{360 °}{n} \\ = \frac{360 °}{10} \\ = 36 ° \end{aligned}

$\begin{aligned} x & = \text{Size of Ext angle of polygon} \\ & = \frac{\text{360}\text{°}}{n} \\ & = \frac{\text{360}\text{°}}{10} \\ & = \text{36}\text{°} \end{aligned}$

We could also calculate the interior angle and then use supplementary rule to find $x$ .

The sum of the interior angles of a polygon is given by

180 ° (n - 2)

$\text{180}\text{°}(n-2)$ .

Therefore the sum of the angles in this polygon is:

\begin{aligned} 180 ° ((10) - 2) & = 180 ° (8) \\ = 1,440 ° \end{aligned}

$\begin{aligned} \text{180}\text{°} ((10) - 2) & = \text{180}\text{°} (8) \\ & = \text{1,440}\text{°} \end{aligned}$

We can use this to work out what each individual interior angle is.

\begin{aligned} interior angle & = 1,440 ° \div 10 \\ = 144 ° \end{aligned}

$\begin{aligned} \text{interior angle} & = \text{1,440}\text{°} \div 10 \\ & = \text{144}\text{°} \end{aligned}$

Finally, we know that the sum of angles on a straight line is $\text{180}\text{°}$ . So

\begin{aligned} exterior angle & = 180 ° - interior angle \\ = 180 ° - 144 ° \\ = 36 ° \end{aligned}

$\begin{aligned} \text{exterior angle} & = \text{180}\text{°} - \text{interior angle} \\ & = \text{180}\text{°} - \text{144}\text{°} \\ & = \text{36}\text{°} \end{aligned}$

So $x = \text{36}\text{°}$

Submit your answer as:

ID is: 1908 Seed is: 6925

Problems with polygons

You are given the following regular polygon with a extension line:

What is the size of exterior angle $x$ for the given polygon?

Answer:

$x =$ °

one-of

type(numeric.abserror(0.01))

2 attempts remaining

STEP: <no title>
[−2 points ⇒ 0 / 2 points left]

Let $n$ be the number of sides. Then $n = \text{9}$ .

\begin{aligned} x & = Size of Ext angle of polygon \\ = \frac{360 °}{n} \\ = \frac{360 °}{9} \\ = 40 ° \end{aligned}

$\begin{aligned} x & = \text{Size of Ext angle of polygon} \\ & = \frac{\text{360}\text{°}}{n} \\ & = \frac{\text{360}\text{°}}{9} \\ & = \text{40}\text{°} \end{aligned}$

We could also calculate the interior angle and then use supplementary rule to find $x$ .

The sum of the interior angles of a polygon is given by

180 ° (n - 2)

$\text{180}\text{°}(n-2)$ .

Therefore the sum of the angles in this polygon is:

\begin{aligned} 180 ° ((9) - 2) & = 180 ° (7) \\ = 1,260 ° \end{aligned}

$\begin{aligned} \text{180}\text{°} ((9) - 2) & = \text{180}\text{°} (7) \\ & = \text{1,260}\text{°} \end{aligned}$

We can use this to work out what each individual interior angle is.

\begin{aligned} interior angle & = 1,260 ° \div 9 \\ = 140 ° \end{aligned}

$\begin{aligned} \text{interior angle} & = \text{1,260}\text{°} \div 9 \\ & = \text{140}\text{°} \end{aligned}$

Finally, we know that the sum of angles on a straight line is $\text{180}\text{°}$ . So

\begin{aligned} exterior angle & = 180 ° - interior angle \\ = 180 ° - 140 ° \\ = 40 ° \end{aligned}

$\begin{aligned} \text{exterior angle} & = \text{180}\text{°} - \text{interior angle} \\ & = \text{180}\text{°} - \text{140}\text{°} \\ & = \text{40}\text{°} \end{aligned}$

So $x = \text{40}\text{°}$

Submit your answer as:

3. Properties of quadrilaterals

ID is: 4282 Seed is: 5364

Apply diagonal properties of a rhombus

In the diagram below, $PQRS$ is a rhombus. Also, $PS =$ $\text{10}$ and $SM =$ $\text{8}$ .

Determine, with reasons, the length of $PR$ .

INSTRUCTION: Answer this question by completing the steps below.

Answer:

P \hat{M} S =

$P\hat MS =$

°

$\text{°}$

∴

$\therefore$

P M =

$PM =$

units

$\text{units}$
Hence,

P R =

$PR =$

units

$\text{units}$

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

What do you need to know about a triangle, before you can use the theorem of Pythagoras? Use the properties that you know about the diagonals of a rhombus.

STEP: Decide on a strategy
[−1 point ⇒ 5 / 6 points left]

$PR = PM + MR$ . In a rhombus, $PM = MR$ . So, if we know $PM$ , we can work out $PR$ . We are going to use the theorem of Pythagoras in $\Delta PMS$ to determine $PM$ . To do this, we will use rhombus diagonal properties to prove that $\Delta PMS$ is right-angled.

STEP: Show that

Δ P M S

$\Delta PMS$ is right-angled
[−1 point ⇒ 4 / 6 points left]

The diagonals of a rhombus intersect at

90 °

$\text{90}\text{°}$ .

$P\hat MS = \text{90}\text{°}$ (diags of rhombus)

STEP: Use the theorem of Pythagoras to determine the length of

P M

$PM$
[−2 points ⇒ 2 / 6 points left]

We have demonstrated that $\Delta PMS$ is right-angled. Now, we can use the theorem of Pythagoras to work out $PM$ .

${\color{blue}{PM^2}} + {\color{red}{SM^2}} = {\color{ForestGreen}{PS^2}}$ (Pythagoras)

\begin{aligned} P M^{2} + 8^{2} = & 10^{2} \\ P M^{2} + 64 = & 100 \\ P M^{2} = & 36 \\ P M = & \sqrt{36} \\ P M = & 6 units \end{aligned}

$\begin{aligned} PM^2 + 8^2 =\: & 10^2 \\ PM^2 + 64 =\: & 100 \\ PM^2 =\: & 36 \\ PM =\: & \sqrt{36} \\ PM =\: & \text{6} \ \text{units} \end{aligned}$

STEP: Apply diagonal properties to determine

P R

$PR$
[−2 points ⇒ 0 / 6 points left]

The diagonals of a rhombus bisect each other.

We know that $PM =$ $\text{6}$ so:

${\color{blue}{PM}} = {\color{magenta}{MR}}$ (diags of rhombus)
$\therefore MR =$ $\text{6}\ \text{units}$

Also, $PR = {\color{blue}{PM}} + {\color{magenta}{MR}}$ (from the diagram)
$\therefore PR =$ $\text{12}\ \text{units}$

Submit your answer as: andandandandand

ID is: 4282 Seed is: 4149

Apply diagonal properties of a rhombus

In the diagram below, $ABCD$ is a rhombus. Also, $\color{magenta}{AD = }$ $\color{magenta}{\text{25}}$ and $\color{teal}{AC = }$ $\color{teal}{\text{30}}$ .

Determine, with reasons, the length of $DE$ .

INSTRUCTION: Answer this question by completing the steps below.

Answer:

A E =

$AE =$

units

$\text{units}$
Also,

A \hat{E} D =

$A\hat ED =$

°

$\text{°}$
Hence,

D E =

$DE =$

units

$\text{units}$

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

What do you need to know about a triangle, before you can use the theorem of Pythagoras? Use the properties that you know about the diagonals of a rhombus.

STEP: Decide on a strategy
[−1 point ⇒ 5 / 6 points left]

We are going to use the theorem of Pythagoras in $\Delta AED$ to determine $AE$ . To do this, we will use rhombus diagonal properties to determine the necessary sides, and to prove that $\Delta AED$ is right-angled.

STEP: Apply diagonal properties to determine

A E

$AE$
[−1 point ⇒ 4 / 6 points left]

The diagonals of a rhombus bisect each other.

$AC = 30$ (given)
And, $AE = EC$ (diags of rhombus)
$\therefore AE =$ $\text{15}\ \text{units}$

STEP: Show that

Δ A E D

$\Delta AED$ is right-angled
[−2 points ⇒ 2 / 6 points left]

The diagonals of a rhombus intersect at

90 °

$\text{90}\text{°}$ .

$A\hat ED = \text{90}\text{°}$ (diags of rhombus)

STEP: Use the theorem of Pythagoras to determine the length of

D E

$DE$
[−2 points ⇒ 0 / 6 points left]

We have shown that $\Delta AED$ is right-angled. Now, we can use the theorem of Pythagoras to work out $DE$ .

${\color{blue}{DE^2}} + {\color{red}{AE^2}} = {\color{ForestGreen}{AD^2}}$ (Pythagoras)

\begin{aligned} D E^{2} + 15^{2} = & 25^{2} \\ D E^{2} + 225 = & 625 \\ D E^{2} = & 400 \\ D E = & \sqrt{400} \\ D E = & 20 units \end{aligned}

$\begin{aligned} DE^2 + 15^2 =\: & 25^2 \\ DE^2 + 225 =\: & 625 \\ DE^2 =\: & 400\\ DE =\: & \sqrt{400}\\ DE =\: & \text{20}\ \text{units} \end{aligned}$

Submit your answer as: andandandandand

ID is: 4282 Seed is: 326

Apply diagonal properties of a rhombus

In the diagram below, $ABCD$ is a rhombus. Also, $AD =$ $\text{13}$ and $AE =$ $\text{5}$ .

Determine, with reasons, the length of $DB$ .

INSTRUCTION: Answer this question by completing the steps below.

Answer:

A \hat{E} D =

$A\hat ED =$

°

$\text{°}$

∴

$\therefore$

D E =

$DE =$

units

$\text{units}$
Hence,

D B =

$DB =$

units

$\text{units}$

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

What do you need to know about a triangle, before you can use the theorem of Pythagoras? Use the properties that you know about the diagonals of a rhombus.

STEP: Decide on a strategy
[−1 point ⇒ 5 / 6 points left]

$DB = DE + EB$ . In a rhombus, $DE = EB$ . So, if we know $DE$ , we can work out $DB$ . We are going to use the theorem of Pythagoras in $\Delta AED$ to determine $DE$ . To do this, we will use rhombus diagonal properties to prove that $\Delta AED$ is right-angled.

STEP: Show that

Δ A E D

$\Delta AED$ is right-angled
[−1 point ⇒ 4 / 6 points left]

The diagonals of a rhombus intersect at

90 °

$\text{90}\text{°}$ .

$A\hat ED = \text{90}\text{°}$ (diags of rhombus)

STEP: Use the theorem of Pythagoras to determine the length of

D E

$DE$
[−2 points ⇒ 2 / 6 points left]

We have demonstrated that $\Delta AED$ is right-angled. Now, we can use the theorem of Pythagoras to work out $DE$ .

${\color{blue}{DE^2}} + {\color{red}{AE^2}} = {\color{ForestGreen}{AD^2}}$ (Pythagoras)

\begin{aligned} D E^{2} + 5^{2} = & 13^{2} \\ D E^{2} + 25 = & 169 \\ D E^{2} = & 144 \\ D E = & \sqrt{144} \\ D E = & 12 units \end{aligned}

$\begin{aligned} DE^2 + 5^2 =\: & 13^2 \\ DE^2 + 25 =\: & 169 \\ DE^2 =\: & 144 \\ DE =\: & \sqrt{144} \\ DE =\: & \text{12} \ \text{units} \end{aligned}$

STEP: Apply diagonal properties to determine

D B

$DB$
[−2 points ⇒ 0 / 6 points left]

The diagonals of a rhombus bisect each other.

We know that $DE =$ $\text{12}$ so:

${\color{blue}{DE}} = {\color{magenta}{EB}}$ (diags of rhombus)
$\therefore EB =$ $\text{12}\ \text{units}$

Also, $DB = {\color{blue}{DE}} + {\color{magenta}{EB}}$ (from the diagram)
$\therefore DB =$ $\text{24}\ \text{units}$

Submit your answer as: andandandandand

ID is: 4214 Seed is: 972

Reasoning about properties of quadrilaterals

Consider the following statement and decide if it is always true, sometimes true, or never true.

In parallelograms, the opposite sides are equal in length.

Answer:

The statement is true.

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Are there any quadrilaterals that make the statement true? Are there any quadrilaterals that make the statement false?

STEP: Think about whether any quadrilaterals make the statement true
[−2 points ⇒ 0 / 2 points left]

Even though it is not part of the definition, the opposite sides of a parallelogram are always the same length.

So, the statement is always true.

Submit your answer as:

ID is: 4214 Seed is: 5435

Reasoning about properties of quadrilaterals

Consider the following statement and decide if it is always true, sometimes true, or never true.

Rectangles have two pairs of opposite sides that are parallel.

Answer:

The statement is true.

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Are there any quadrilaterals that make the statement true? Are there any quadrilaterals that make the statement false?

STEP: Think about whether any quadrilaterals make the statement true
[−2 points ⇒ 0 / 2 points left]

The opposite sides of a rectangle are always parallel, because a rectangle is a special type of parallelogram.

So, the statement is always true.

Submit your answer as:

ID is: 4214 Seed is: 1428

Reasoning about properties of quadrilaterals

Consider the following statement and decide if it is always true, sometimes true, or never true.

Squares have two pairs of opposite sides that are parallel.

Answer:

The statement is true.

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Are there any quadrilaterals that make the statement true? Are there any quadrilaterals that make the statement false?

STEP: Think about whether any quadrilaterals make the statement true
[−2 points ⇒ 0 / 2 points left]

The opposite sides of a square are always parallel, because a square is a special type of parallelogram.

So, the statement is always true.

Submit your answer as:

ID is: 4250 Seed is: 5688

Identify all diagonal properties

Think about the diagonals of a rhombus.

Select whether each of the following statements is true or false for all rhombuses.

Answer:

The diagonals bisect each other.
Both diagonals are the same length.
The diagonals intersect at $\text{90}\text{°}$ .
The diagonals bisect the corner angles.

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Draw a standard rhombus and compare the diagram to each diagonal property.

STEP: Identify the diagonal properties
[−4 points ⇒ 0 / 4 points left]

Rhombuses have three diagonal properties that we can prove.

The diagonals of a kite intersect at $\text{90}\text{°}$ . A rhombus is a special type of kite, so the diagonals of a rhombus must also intersect at $\text{90}\text{°}$ .

The diagonals of a parallelogram bisect each other. A rhombus is a special type of parallelogram, so the diagonals of a rhombus must also bisect each other.

When we draw in the diagonals, we create several pairs of congruent triangles. In particular $\Delta PTS$ , $\Delta RTS$ , $\Delta RTQ$ and $\Delta PTQ$ are all congruent to each other. (Think about how you would prove this!)

This means that:

$\color{orange}{P\hat ST = R\hat ST}$ ,
$\color{teal}{S\hat RT = Q\hat RT}$ ,
$\color{blue}{R\hat QT = P\hat QT}$ , and
$\color{magenta}{Q\hat PT = S\hat PT}$ .

The corner angles are all cut into two equal parts by the diagonals, so the diagonals bisect the corner angles.

These are the only properties that we can prove for certain about the diagonals of all rhombuses. The following table shows the correct answers:

Diagonal property	Definitely true for any rhombus?
The diagonals bisect each other.	✔ True
Both diagonals are the same length.	✘ False
The diagonals intersect at $\text{90}\text{°}$ .	✔ True
The diagonals bisect the corner angles.	✔ True

Submit your answer as: andandand

ID is: 4250 Seed is: 5741

Identify all diagonal properties

Think about the diagonals of a square.

Select whether each of the following statements is true or false for all squares.

Answer:

The diagonals bisect each other.
Both diagonals are the same length.
The diagonals intersect at $\text{90}\text{°}$ .
The diagonals bisect the corner angles.

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Draw a standard square and compare the diagram to each diagonal property.

STEP: Identify the diagonal properties
[−4 points ⇒ 0 / 4 points left]

A square is a special type of rectangle and a special type of rhombus. So it will have all of the diagonal properties of rectangles and of rhombuses.

The following table shows the correct answers:

Diagonal property	Definitely true for any square?
The diagonals bisect each other.	✔ True
Both diagonals are the same length.	✔ True
The diagonals intersect at $\text{90}\text{°}$ .	✔ True
The diagonals bisect the corner angles.	✔ True

Submit your answer as: andandand

ID is: 4250 Seed is: 3628

Identify all diagonal properties

Think about the diagonals of a kite.

Select whether each of the following statements is true or false for all kites.

Answer:

The diagonals bisect each other.
Both diagonals are the same length.
The diagonals intersect at $\text{90}\text{°}$ .
The diagonals bisect the corner angles.

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Draw a standard kite and compare the diagram to each diagonal property.

STEP: Identify the diagonal properties
[−4 points ⇒ 0 / 4 points left]

When we draw in the diagonals, we create several pairs of congruent triangles, including $\Delta ABE \equiv \Delta ADE$ . (This is a bit harder to prove - but you should still try!) This means that $E_1$ and $E_2$ are equal. We also know that $E_1$ and $E_2$ add up to $\text{180}\text{°}$ . So, they must both be $\text{90}\text{°}$ . So, the diagonals intersect at $\text{90}\text{°}$ .

NOTE: One diagonal of a kite,

B D

$BD$ , is bisected by the other. But

A C

$AC$ is not bisected. So, we do not say that the diagonals bisect each other in a kite.

This is the only property that we can prove for certain about the diagonals of all kites. The following table shows the correct answers:

Diagonal property	Definitely true for any kite?
The diagonals bisect each other.	✘ False
Both diagonals are the same length.	✘ False
The diagonals intersect at $\text{90}\text{°}$ .	✔ True
The diagonals bisect the corner angles.	✘ False

Submit your answer as: andandand

ID is: 1654 Seed is: 6367

Congruent triangles

Have a look at the following triangles, which are drawn to scale:

Answer the two questions below.
Answer:
Are these triangles congruent?
If yes, select the reason. If no, select 'not congruent.'
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

The word "congruent" means "same shape". Look carefully at the triangles and the given information. Can you prove that these triangles are congruent? If so, remember to provide a correct reason.

STEP: Examine the picture to see if the triangles are identical
[−2 points ⇒ 0 / 2 points left]

The word "congruent" means "identical." If the triangles have exactly the same shape, then they are congruent.

There are four different ways to determine if two triangles are congruent.
1. Right angle, hypotenuse, side (RHS): the two triangles are right-angled triangles, the hypotenuse of one triangle is the same length as the hypotenuse of the other triangle and one of the other pairs of corresponding sides are equal in length.
2. Side, side, side (SSS): all three pairs of corresponding sides in the two triangles have equal lengths.
3. Side, angle, side (SAS): two pairs of corresponding sides and the included angle are equal.
4. Angle, angle, side (AAS): two pairs of corresponding angles are equal and one pair of corresponding sides have equal lengths.
Note that the two pairs of sides are equal, as indicated by the $m$ and $n$ . In addition, the angle between those two sides are marked as equal.

Therefore, these two triangles are congruent, and the reason is SAS.

Submit your answer as: and
Which of the following options correctly completes the statement below?

Answer: $\Delta CAB \equiv \Delta$

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]
When shapes are congruent, you must label them in the order of their corresponding angles. "Corresponding" means "matching." Look at the triangles: does $\hat{A}$ correspond to angle $\hat{D}, \hat{E}$ , or $\hat{F}$ ?

STEP: Compare the triangles to complete the statement
[−1 point ⇒ 0 / 1 points left]

When we name a pair of congruent triangles we need to pay careful attention to the order of the angles and which sides are equal: the triangles must be named according to their corresponding angles and sides. "Corresponding" means "matching."

The first triangle can be named in any order - we are given $\Delta CAB$ . The second triangle needs to be named so that we can read off the corresponding angles and sides: we need to ensure that the corresponding sides and angles match up.

For example, vertex $C$ corresponds to vertex $F$ . Therefore, the answer must start with $F$ . Now think about which vertex corresponds to vertex $A$ to figure out which letter should be second in the answer.

The correct choice to complete the statement is $\Delta FDE$ .

Submit your answer as:

ID is: 1654 Seed is: 2637

Congruent triangles

Have a look at the following triangles, which are drawn to scale:

Answer the two questions below.
Answer:
Are these triangles congruent?
If yes, select the reason. If no, select 'not congruent.'
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

The word "congruent" means "same shape". Look carefully at the triangles and the given information. Can you prove that these triangles are congruent? If so, remember to provide a correct reason.

STEP: Examine the picture to see if the triangles are identical
[−2 points ⇒ 0 / 2 points left]

The word "congruent" means "identical." If the triangles have exactly the same shape, then they are congruent.

There are four different ways to determine if two triangles are congruent.
1. Right angle, hypotenuse, side (RHS): the two triangles are right-angled triangles, the hypotenuse of one triangle is the same length as the hypotenuse of the other triangle and one of the other pairs of corresponding sides are equal in length.
2. Side, side, side (SSS): all three pairs of corresponding sides in the two triangles have equal lengths.
3. Side, angle, side (SAS): two pairs of corresponding sides and the included angle are equal.
4. Angle, angle, side (AAS): two pairs of corresponding angles are equal and one pair of corresponding sides have equal lengths.
We are given information about two angles and one side: there are two pairs of corresponding angles that are equal (indicated by the dots and the stars), and one pair of sides which are the same length.

Therefore, these two triangles are congruent, and the reason is AAS.

Submit your answer as: and
Which of the following options correctly completes the statement below?

Answer: $\Delta VUW \equiv \Delta$

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]
When shapes are congruent, you must label them in the order of their corresponding angles. "Corresponding" means "matching." Look at the triangles: does $\hat{U}$ correspond to angle $\hat{X}, \hat{Y}$ , or $\hat{Z}$ ?

STEP: Compare the triangles to complete the statement
[−1 point ⇒ 0 / 1 points left]

When we name a pair of congruent triangles we need to pay careful attention to the order of the angles and which sides are equal: the triangles must be named according to their corresponding angles and sides. "Corresponding" means "matching."

The first triangle can be named in any order - we are given $\Delta VUW$ . The second triangle needs to be named so that we can read off the corresponding angles and sides: we need to ensure that the corresponding sides and angles match up.

For example, vertex $V$ corresponds to vertex $Y$ . Therefore, the answer must start with $Y$ . Now think about which vertex corresponds to vertex $U$ to figure out which letter should be second in the answer.

The correct choice to complete the statement is $\Delta YXZ$ .

Submit your answer as:

ID is: 1654 Seed is: 4741

Congruent triangles

Have a look at the following triangles, which are drawn to scale:

Answer the two questions below.
Answer:
Are these triangles congruent?
If yes, select the reason. If no, select 'not congruent.'
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

The word "congruent" means "same shape". Look carefully at the triangles and the given information. Can you prove that these triangles are congruent? If so, remember to provide a correct reason.

STEP: Examine the picture to see if the triangles are identical
[−2 points ⇒ 0 / 2 points left]

The word "congruent" means "identical." If the triangles have exactly the same shape, then they are congruent.

There are four different ways to determine if two triangles are congruent.
1. Right angle, hypotenuse, side (RHS): the two triangles are right-angled triangles, the hypotenuse of one triangle is the same length as the hypotenuse of the other triangle and one of the other pairs of corresponding sides are equal in length.
2. Side, side, side (SSS): all three pairs of corresponding sides in the two triangles have equal lengths.
3. Side, angle, side (SAS): two pairs of corresponding sides and the included angle are equal.
4. Angle, angle, side (AAS): two pairs of corresponding angles are equal and one pair of corresponding sides have equal lengths.
The sides of both triangles are labelled with $m$ , $n$ and $p$ . This means that there are three pairs of corresponding and equal sides.

Therefore, these two triangles are congruent, and the reason is SSS.

Submit your answer as: and
Which of the following options correctly completes the statement below?

Answer: $\Delta VUW \equiv \Delta$

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]
When shapes are congruent, you must label them in the order of their corresponding angles. "Corresponding" means "matching." Look at the triangles: does $\hat{U}$ correspond to angle $\hat{X}, \hat{Y}$ , or $\hat{Z}$ ?

STEP: Compare the triangles to complete the statement
[−1 point ⇒ 0 / 1 points left]

When we name a pair of congruent triangles we need to pay careful attention to the order of the angles and which sides are equal: the triangles must be named according to their corresponding angles and sides. "Corresponding" means "matching."

The first triangle can be named in any order - we are given $\Delta VUW$ . The second triangle needs to be named so that we can read off the corresponding angles and sides: we need to ensure that the corresponding sides and angles match up.

For example, vertex $V$ corresponds to vertex $Y$ . Therefore, the answer must start with $Y$ . Now think about which vertex corresponds to vertex $U$ to figure out which letter should be second in the answer.

The correct choice to complete the statement is $\Delta YXZ$ .

Submit your answer as:

ID is: 1865 Seed is: 636

Calculate angles between parallel lines

The diagram represents transversal line $FG$ that intersects with the straight lines $BC$ and $DE$ respectively. $BC \parallel$ $DE$ .

The intersecting lines create different angles which are labeled for you. Carefully study the diagram and answer the questions that follow.

Answer:

What is the value of $x$ ? ${°}$
What is the value of $y$ ? ${°}$
What is the value of $z$ ? ${°}$

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

STEP: Determine the unknown angles
[−3 points ⇒ 0 / 3 points left]

If the two lines are parallel, the four angles around the first intersection are the same as the four angles around the second intersection. We can use this, and the fact that angles on a straight line add up to $180°$ , to determine the unknown angles.

Therefore:

${\color{red}{x = 82 {°}}}$
${\color{blue}{y = 82 {°}}}$
${\color{darkgreen}{z = 98 {°}}}$

Submit your answer as: andand

ID is: 1865 Seed is: 4969

Calculate angles between parallel lines

The diagram represents transversal line $LM$ that intersects with the straight lines $GH$ and $JK$ respectively. $GH \parallel$ $JK$ .

The intersecting lines create different angles which are labeled for you. Carefully study the diagram and answer the questions that follow.

Answer:

What is the value of $x$ ? ${°}$
What is the value of $y$ ? ${°}$
What is the value of $z$ ? ${°}$

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

STEP: Determine the unknown angles
[−3 points ⇒ 0 / 3 points left]

Therefore:

${\color{red}{x = 78 {°}}}$
${\color{blue}{y = 78 {°}}}$
${\color{darkgreen}{z = 102 {°}}}$

Submit your answer as: andand

ID is: 1865 Seed is: 1784

Calculate angles between parallel lines

The diagram represents transversal line $EF$ that intersects with the straight lines $AB$ and $CD$ respectively. $AB \parallel$ $CD$ .

The intersecting lines create different angles which are labeled for you. Carefully study the diagram and answer the questions that follow.

Answer:

What is the value of $x$ ? ${°}$
What is the value of $y$ ? ${°}$
What is the value of $z$ ? ${°}$

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

STEP: Determine the unknown angles
[−3 points ⇒ 0 / 3 points left]

Therefore:

${\color{red}{x = 81 {°}}}$
${\color{blue}{y = 81 {°}}}$
${\color{darkgreen}{z = 99 {°}}}$

Submit your answer as: andand

ID is: 1670 Seed is: 6143

Proof: is the quadrilateral a parallelogram?

The figure below shows quadrilateral $MNOP$ with $MP \parallel NO$ and $MP = NO$ . Diagonal $NP$ is shown with a dashed line. $\angle M = y$ and $\angle O = 74°$ ; $\angle MPN = x$ and $\angle NPO = 57°$ .

The steps and reasons below prove that $MNOP$ is a parallelogram. But some steps and reasons are incomplete. Choose the correct steps and reasons from the drop down boxes to complete the proof.

Answer:

$\quad$ STEPS $\quad$	REASONS
	Alternate interior angles are equal if $MP \parallel NO$
$NP = PN$	Common side (in triangles $MNP$ & $OPN$ )
$MP = NO$	Given
$\Delta MNP = \Delta OPN$	Side-Angle-Side for congruent triangles (SAS)

$MN \parallel PO$	$\angle MNP$ & $\angle OPN$ are equal alternate interior angles
$MNOP$ is a parallelogram

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

We already know that side $MP$ is parallel to side $NO$ . The proof above shows that the other two sides are also parallel, based on the information given about the angles. Work your way down the proof, one step at a time, to choose the correct steps and reasons.

STEP: Work your way down the proof, choosing the answers along the way
[−4 points ⇒ 0 / 4 points left]

The overall strategy of the proof in this question is to show that the sides $MN$ and $PO$ are parallel. If we can show that, we can conclude that the quadrilateral is a parallelogram because we already know that $MP$ is parallel to $NO$ . But it takes six steps to first prove that $MN$ and $PO$ are parallel.

The proof starts by using the parallel sides and the diagonal. The parallel lines and the diagonal invite us to use transversal geometry: alternating interior angles, corresponding angles, cointerior angles, and so on. So we can show that two of the angles are equal. Then the proof proceeds to show (in step $\text{4}$ ) that the triangles $MNP$ and $OPN$ are congruent (which means that the triangles are identical).

The complete proof is as follows:

$\quad$ Steps $\quad$	Reasons
$\angle MPN = \angle PNO$	Alternate interior angles are equal if $MP \parallel NO$
$NP = PN$	Common side (in triangles $MNP$ & $OPN$ )
$MP = NO$	Given
$\Delta MNP = \Delta OPN$	Side-Angle-Side for congruent triangles (SAS)
$\angle MNP = \angle OPN$	Corresponding angles in congruent triangles
$MN \parallel PO$	$\angle MNP$ & $\angle OPN$ are equal alternate interior angles
$MNOP$ is a parallelogram	Definition of a parallelogram (opposite sides parallel)

Submit your answer as: andandand

Determine the value of $y$ . The diagram is repeated here, but now with both sides labelled as parallel (which we know from Question 1).

Answer: $y =$ °
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Use the fact that opposite angles in a parallelogram are equal.

STEP: Use the fact that opposite angles in a parallelogram are equal
[−1 point ⇒ 0 / 1 points left]

$MNOP$ is a parallelogram. Opposite angles of a parallelogram are equal. So $\angle M$ is equal to $\angle O$ .

Therefore, $y = 74°$ .

Submit your answer as:
Determine the value of $x$ .

Answer: $x =$ °
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by writing an equation for the angles in triangle $MNP$ . The sum of those angles, which include $x$ , must be $180°$ .

STEP: Set up an equation and solve for $x$
[−2 points ⇒ 0 / 2 points left]

We can solve this problem using the fact that angle $x$ is inside of a triangle $MNP$ .

The sum of angles in a triangle is $180°$ , so based on the figure we can write:

$\begin{array}{c}\angle M + \angle MPN + \angle MNP = 180°\end{array}$

From Question 2 we know that $y = 74°$ (this is labelled in the diagram above). And because of alternate interior angles we also know that $\angle MNP = 57°$ . Putting all of this together:

$\begin{aligned} \angle M + \angle MPN + \angle MNP & = 180° \\ 74° + x + 57° & = 180° \\ x & = 180° - 74° - 57° \\ x & = 49° \end{aligned}$

NOTE: You can also solve this question using the fact that the angles in a quadrliateral have a sum of $360°$ . Using the fact that opposite angles in a parallelogram are equal, the equation would be:
$\begin{array}{c}74° + 74° + 2(57° + x) = 360°\end{array}$
If you solve this you will also get the answer $x = 49°$ .

The measure of angle $x$ is $49°$ .

Submit your answer as:

ID is: 1670 Seed is: 5538

Proof: is the quadrilateral a parallelogram?

The figure below shows quadrilateral $ABCD$ with $AD \parallel BC$ and $AD = BC$ . Diagonal $BD$ is shown with a dashed line. $\angle A = y$ and $\angle C = 76°$ ; $\angle ADB = 45°$ and $\angle BDC = x$ .

The steps and reasons below prove that $ABCD$ is a parallelogram. But some steps and reasons are incomplete. Choose the correct steps and reasons from the drop down boxes to complete the proof.

Answer:

$\quad$ STEPS $\quad$	REASONS
	Alternate interior angles are equal if $AD \parallel BC$
	Common side (in triangles $ABD$ & $CDB$ )
$AD = BC$
$\Delta ABD = \Delta CDB$
$\angle ABD = \angle CDB$	Corresponding angles in congruent triangles
$AB \parallel DC$	$\angle ABD$ & $\angle CDB$ are equal alternate interior angles
$ABCD$ is a parallelogram	Definition of a parallelogram (opposite sides parallel)

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

We already know that side $AD$ is parallel to side $BC$ . The proof above shows that the other two sides are also parallel, based on the information given about the angles. Work your way down the proof, one step at a time, to choose the correct steps and reasons.

STEP: Work your way down the proof, choosing the answers along the way
[−4 points ⇒ 0 / 4 points left]

The overall strategy of the proof in this question is to show that the sides $AB$ and $DC$ are parallel. If we can show that, we can conclude that the quadrilateral is a parallelogram because we already know that $AD$ is parallel to $BC$ . But it takes six steps to first prove that $AB$ and $DC$ are parallel.

The proof starts by using the parallel sides and the diagonal. The parallel lines and the diagonal invite us to use transversal geometry: alternating interior angles, corresponding angles, cointerior angles, and so on. So we can show that two of the angles are equal. Then the proof proceeds to show (in step $\text{4}$ ) that the triangles $ABD$ and $CDB$ are congruent (which means that the triangles are identical).

The complete proof is as follows:

$\quad$ Steps $\quad$	Reasons
$\angle ADB = \angle DBC$	Alternate interior angles are equal if $AD \parallel BC$
$BD = DB$	Common side (in triangles $ABD$ & $CDB$ )
$AD = BC$	Given
$\Delta ABD = \Delta CDB$	Side-Angle-Side for congruent triangles (SAS)
$\angle ABD = \angle CDB$	Corresponding angles in congruent triangles
$AB \parallel DC$	$\angle ABD$ & $\angle CDB$ are equal alternate interior angles
$ABCD$ is a parallelogram	Definition of a parallelogram (opposite sides parallel)

Submit your answer as: andandand

Determine the value of $y$ . The diagram is repeated here, but now with both sides labelled as parallel (which we know from Question 1).

Answer: $y =$ °
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Use the fact that opposite angles in a parallelogram are equal.

STEP: Use the fact that opposite angles in a parallelogram are equal
[−1 point ⇒ 0 / 1 points left]

$ABCD$ is a parallelogram. Opposite angles of a parallelogram are equal. So $\angle A$ is equal to $\angle C$ .

Therefore, $y = 76°$ .

Submit your answer as:
Determine the value of $x$ .

Answer: $x =$ °
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by writing an equation for the angles in triangle $CBD$ . The sum of those angles, which include $x$ , must be $180°$ .

STEP: Set up an equation and solve for $x$
[−2 points ⇒ 0 / 2 points left]

We can solve this problem using the fact that angle $x$ is inside of a triangle $CBD$ .

The sum of angles in a triangle is $180°$ , so based on the figure we can write:

$\begin{array}{c}\angle C + \angle CDB + \angle CBD = 180°\end{array}$

From Question 2 we know that $y = 76°$ (this is labelled in the diagram above). And because of alternate interior angles we also know that $\angle CBD = 45°$ . Putting all of this together:

$\begin{aligned} \angle C + \angle CDB + \angle CBD & = 180° \\ 76° + x + 45° & = 180° \\ x & = 180° - 76° - 45° \\ x & = 59° \end{aligned}$

NOTE: You can also solve this question using the fact that the angles in a quadrliateral have a sum of $360°$ . Using the fact that opposite angles in a parallelogram are equal, the equation would be:
$\begin{array}{c}76° + 76° + 2(45° + x) = 360°\end{array}$
If you solve this you will also get the answer $x = 59°$ .

The measure of angle $x$ is $59°$ .

Submit your answer as:

ID is: 1670 Seed is: 685

Proof: is the quadrilateral a parallelogram?

The figure below shows quadrilateral $MNOP$ with $MP \parallel NO$ and $MP = NO$ . Diagonal $NP$ is shown with a dashed line. $\angle M = y$ and $\angle O = 76°$ ; $\angle MPN = 67°$ and $\angle NPO = x$ .

The steps and reasons below prove that $MNOP$ is a parallelogram. But some steps and reasons are incomplete. Choose the correct steps and reasons from the drop down boxes to complete the proof.

Answer:

$\quad$ STEPS $\quad$	REASONS

$NP = PN$
$MP = NO$	Given
$\Delta MNP = \Delta OPN$	Side-Angle-Side for congruent triangles (SAS)
	Corresponding angles in congruent triangles
$MN \parallel PO$	$\angle MNP$ & $\angle OPN$ are equal alternate interior angles
$MNOP$ is a parallelogram	Definition of a parallelogram (opposite sides parallel)

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

STEP: Work your way down the proof, choosing the answers along the way
[−4 points ⇒ 0 / 4 points left]

The complete proof is as follows:

$\quad$ Steps $\quad$	Reasons
$\angle MPN = \angle PNO$	Alternate interior angles are equal if $MP \parallel NO$
$NP = PN$	Common side (in triangles $MNP$ & $OPN$ )
$MP = NO$	Given
$\Delta MNP = \Delta OPN$	Side-Angle-Side for congruent triangles (SAS)
$\angle MNP = \angle OPN$	Corresponding angles in congruent triangles
$MN \parallel PO$	$\angle MNP$ & $\angle OPN$ are equal alternate interior angles
$MNOP$ is a parallelogram	Definition of a parallelogram (opposite sides parallel)

Submit your answer as: andandand

Determine the value of $y$ . The diagram is repeated here, but now with both sides labelled as parallel (which we know from Question 1).

Answer: $y =$ °
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Use the fact that opposite angles in a parallelogram are equal.

STEP: Use the fact that opposite angles in a parallelogram are equal
[−1 point ⇒ 0 / 1 points left]

$MNOP$ is a parallelogram. Opposite angles of a parallelogram are equal. So $\angle M$ is equal to $\angle O$ .

Therefore, $y = 76°$ .

Submit your answer as:
Determine the value of $x$ .

Answer: $x =$ °
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by writing an equation for the angles in triangle $ONP$ . The sum of those angles, which include $x$ , must be $180°$ .

STEP: Set up an equation and solve for $x$
[−2 points ⇒ 0 / 2 points left]

We can solve this problem using the fact that angle $x$ is inside of a triangle $ONP$ .

The sum of angles in a triangle is $180°$ , so based on the figure we can write:

$\begin{array}{c}\angle O + \angle OPN + \angle ONP = 180°\end{array}$

From Question 2 we know that $y = 76°$ (this is labelled in the diagram above). And because of alternate interior angles we also know that $\angle ONP = 67°$ . Putting all of this together:

$\begin{aligned} \angle O + \angle OPN + \angle ONP & = 180° \\ 76° + x + 67° & = 180° \\ x & = 180° - 76° - 67° \\ x & = 37° \end{aligned}$

NOTE: You can also solve this question using the fact that the angles in a quadrliateral have a sum of $360°$ . Using the fact that opposite angles in a parallelogram are equal, the equation would be:
$\begin{array}{c}76° + 76° + 2(67° + x) = 360°\end{array}$
If you solve this you will also get the answer $x = 37°$ .

The measure of angle $x$ is $37°$ .

Submit your answer as:

ID is: 4251 Seed is: 4170

Choose one correct diagonal property

$PQRS$ below is a parallelogram.

Answer:

Which of the following statements is definitely true about the diagonals of $PQRS$ ?

AThe diagonals bisect the corner angles.
BBoth diagonals are equal in length.
CThe diagonals are perpendicular to each other.
DThe diagonals bisect each other.

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Diagonals go from one corner of the parallelogram to the opposite corner.
Bisect means to cut into two equal parts.
Perpendicular means that the lines meet at $\text{90}\text{°}$ .

STEP: Identify the diagonal properties
[−2 points ⇒ 0 / 2 points left]

When we draw in the diagonals, we create several pairs of congruent triangles. In particular, $\Delta PTQ \equiv \Delta RTS$ . (Think about how you would prove this!) So, $\color{blue}{PT = RT}$ and $\color{magenta}{TS = TQ}$ . In other words, the diagonals bisect each other.

NOTE: Bisect means to cut into two equal parts.

This is the only property that we can prove for certain about the diagonals of all parallelograms.

Diagonal property	Definitely true for any parallelogram?
The diagonals bisect each other.	✔
Both diagonals are the same length.	✘
The diagonals intersect at $\text{90}\text{°}$ .	✘
The diagonals bisect the corner angles.	✘

Only D is definitely true for all parallelograms.

Submit your answer as:

DThe diagonals bisect each other.

ID is: 4251 Seed is: 7788

Choose one correct diagonal property

$ABCD$ below is a parallelogram.

Answer:

Which of the following statements is definitely true about the diagonals of $ABCD$ ?

ABoth diagonals are equal in length.
BThe diagonals bisect each other.
CThe diagonals bisect the corner angles.
DThe diagonals intersect at $\text{90}\text{°}$ .

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Diagonals go from one corner of the parallelogram to the opposite corner.
Bisect means to cut into two equal parts.
Perpendicular means that the lines meet at $\text{90}\text{°}$ .

STEP: Identify the diagonal properties
[−2 points ⇒ 0 / 2 points left]

When we draw in the diagonals, we create several pairs of congruent triangles. In particular, $\Delta AEB \equiv \Delta CED$ . (Think about how you would prove this!) So, $\color{blue}{AE = CE}$ and $\color{magenta}{ED = EB}$ . In other words, the diagonals bisect each other.

NOTE: Bisect means to cut into two equal parts.

This is the only property that we can prove for certain about the diagonals of all parallelograms.

Diagonal property	Definitely true for any parallelogram?
The diagonals bisect each other.	✔
Both diagonals are the same length.	✘
The diagonals intersect at $\text{90}\text{°}$ .	✘
The diagonals bisect the corner angles.	✘

Only B is definitely true for all parallelograms.

Submit your answer as:

BThe diagonals bisect each other.

ID is: 4251 Seed is: 4418

Choose one correct diagonal property

$PQRS$ below is a kite.

Answer:

Which of the following statements is definitely true about the diagonals of $PQRS$ ?

AThe diagonals intersect at $\text{90}\text{°}$ .
BThe diagonals bisect each other.
CThe diagonals bisect the corner angles.
DAll of the above.

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Diagonals go from one corner of the kite to the opposite corner.
Bisect means to cut into two equal parts.
Perpendicular means that the lines meet at $\text{90}\text{°}$ .

STEP: Identify the diagonal properties
[−2 points ⇒ 0 / 2 points left]

When we draw in the diagonals, we create several pairs of congruent triangles, including $\Delta PQT \equiv \Delta PST$ . (This is a bit harder to prove - but you should still try!) This means that $T_1$ and $T_2$ are equal. We also know $T_1$ and $T_2$ add up to $\text{180}\text{°}$ , so they must both be $\text{90}\text{°}$ . This means the diagonals intersect at $\text{90}\text{°}$ .

This is the only property that we can prove for certain about the diagonals of all kites.

Diagonal property	Definitely true for any kite?
The diagonals bisect each other.	✘
Both diagonals are the same length.	✘
The diagonals intersect at $\text{90}\text{°}$ .	✔
The diagonals bisect the corner angles.	✘

Only A is definitely true for all kites.

NOTE: One diagonal of a kite,

Q S

$QS$ , is bisected by the other. But

P R

$PR$ is not bisected. So, we do not say that the diagonals bisect each other in a kite.

Submit your answer as:

AThe diagonals intersect at

90 °

$\text{90}\text{°}$ .

ID is: 4253 Seed is: 4321

Apply diagonal properties

In the diagram below, $A\hat BE = \text{21}\text{°}$ and $A \hat DE = \text{102}\text{°}$ .

Answer the following questions:

What type of quadrilateral is $ABCD$ ? (Give the most specific name.)
Determine the size of $E\hat DC$ .

INSTRUCTION: You do not need to give any reasons for your answers in this question.

Answer:

$ABCD$ is a .
$E\hat DC =$ $\text{°}$

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Think about the diagonal properties of $ABCD$ . How can these help you to determine $E\hat DC$ ?

STEP: Identify

A B C D

$ABCD$
[−1 point ⇒ 2 / 3 points left]

The most specific name for $ABCD$ is a parallelogram. Even though $ABCD$ is a type of trapezium (for example), this is not the most specific name for $ABCD$ .

TIP: When asked to give the name of a quadrilateral, always give the most specific name.

STEP: Apply diagonal properties of a parallelogram to determine

E \hat{D} C

$E\hat DC$
[−2 points ⇒ 0 / 3 points left]

Remember the diagonal properties of a parallelogram:

Diagonal property	Definitely true for a parallelogram?
The diagonals bisect each other.	✔
Both diagonals are the same length.	✘
The diagonals intersect at $\text{90}\text{°}$ .	✘
The diagonals bisect the corner angles.	✘

The diagonals of a parallelogram do not necessarily bisect the corner angles, so we cannot assume that $\color{blue}{E\hat DC}$ is equal to $A\hat DE$ . In fact, we must use the fact that alternate angles on parallel lines are equal to deduce that $\color{blue}{E\hat DC}$ $=$ $\color{red}{A\hat BE}$ $=$ $\text{21}\text{°}$ .

Submit your answer as: and

ID is: 4253 Seed is: 5439

Apply diagonal properties

In the diagram below, $PT = 13$ and $P\hat RS = \text{30}\text{°}$ .

Answer the following questions:

What type of quadrilateral is $PQRS$ ? (Give the most specific name.)
Determine the size of $P\hat RQ$ .

INSTRUCTION: You do not need to give any reasons for your answers in this question.

Answer:

$PQRS$ is a .
$P\hat RQ =$ $\text{°}$

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Think about the diagonal properties of $PQRS$ . How can these help you to determine $P\hat RQ$ ?

STEP: Identify

P Q R S

$PQRS$
[−1 point ⇒ 2 / 3 points left]

The most specific name for $PQRS$ is a rectangle. Even though $PQRS$ is a type of parallelogram (for example), this is not the most specific name for $PQRS$ .

TIP: When asked to give the name of a quadrilateral, always give the most specific name.

STEP: Apply diagonal properties of a rectangle to determine

P \hat{R} Q

$P\hat RQ$
[−2 points ⇒ 0 / 3 points left]

Remember the diagonal properties of a rectangle:

Diagonal property	Definitely true for a rectangle?
The diagonals bisect each other.	✔
Both diagonals are the same length.	✔
The diagonals intersect at $\text{90}\text{°}$ .	✘
The diagonals bisect the corner angles.	✘

The diagonals of a rectangle do not necessarily bisect the corner angles. So, we cannot assume that $\color{blue}{P\hat RQ}$ $=$ $\color{red}{P\hat RS}$ . Instead, we must use the fact that $Q\hat RS = \text{90}\text{°}$ :

$Q\hat RS = \text{90}\text{°}$ and $\color{red}{P\hat RS}$ $= \text{30}\text{°}$ $\text{(both given)}$
$\text{Also,}\: {\color{blue}{P\hat RQ}} + {\color{red}{P\hat RS}} = Q\hat RS$ $\text{(from the diagram)}$

\begin{aligned} ∴ P \hat{R} Q + 30 ° = 90 ° \\ ∴ P \hat{R} Q = 60 ° \end{aligned}

$\begin{aligned} &\therefore {\color{blue}{P\hat RQ}} + \text{30}\text{°} = \text{90}\text{°} \\ &\therefore {\color{blue}{P\hat RQ}} = \text{60}\text{°} \end{aligned}$

Submit your answer as: and

ID is: 4253 Seed is: 8698

Apply diagonal properties

In the diagram below, $T\hat SR = \text{32}\text{°}$ and $T\hat PS = \text{58}\text{°}$ .

Answer the following questions:

What type of quadrilateral is $PQRS$ ? (Give the most specific name.)
Determine the size of $P\hat TS$ .

INSTRUCTION: You do not need to give any reasons for your answers in this question.

Answer:

$PQRS$ is a .
$P\hat TS =$ $\text{°}$

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Think about the diagonal properties of $PQRS$ . How can these help you to determine $P\hat TS$ ?

STEP: Identify

P Q R S

$PQRS$
[−1 point ⇒ 2 / 3 points left]

The most specific name for $PQRS$ is a rhombus. Even though $PQRS$ is a type of parallelogram (for example), this is not the most specific name for $PQRS$ .

TIP: When asked to give the name of a quadrilateral, always give the most specific name.

STEP: Apply diagonal properties of a rhombus to determine

P \hat{T} S

$P\hat TS$
[−2 points ⇒ 0 / 3 points left]

Remember the diagonal properties of a rhombus:

Diagonal property	Definitely true for a rhombus?
The diagonals bisect each other.	✔
Both diagonals are the same length.	✘
The diagonals intersect at $\text{90}\text{°}$ .	✔
The diagonals bisect the corner angles.	✔

The diagonals of a rhombus intersect at $\text{90}\text{°}$ . Therefore, $\color{red}{P\hat TS}$ $=$ $\text{90}\text{°}$ .

Submit your answer as: and

ID is: 1858 Seed is: 6807

Congruent triangles

In the diagram below, $\Delta MNP \equiv \Delta MQP$ . Also, $MP \perp QN$ while $NP = 10$ and $MP = 8$ .

Calculate the value of $x$ .
Determine the length of $QN$ .

Answer:

$x =$ $\text{units}$
$QN =$ $\text{units}$

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

The triangles are congruent. That means they are identical. Start by using this fact to label all the sides which are not already labelled.

STEP: Label everything you can in the figure
[−1 point ⇒ 3 / 4 points left]

The first question asks us to determine the length of the side of the diagram labelled $x$ . The $x$ is attached to side $MQ$ .

Based on the labels in the diagram, we do not know enough to calculate $x$ immediately. However, the question states that the two triangles are congruent. That means they are identical. We can use that information to label all the sides which are not already labelled. The question also states that $MP \perp QN$ , which means that there are two right angles. We should label those right angles too.

Now we can see two right-angled triangles. And we can see which sides are equal to each other. Both of the triangles include a side labelled $x$ .

STEP: Use the theorem of Pythagoras to find the value of

x

$x$
[−2 points ⇒ 1 / 4 points left]

Now calculate $x$ using the theorem of Pythagoras. We will focus on triangle $MQP$ , but you can use either one, because they are congruent.

\begin{aligned} In & Δ M Q P : \\ Q P & = 10 (Δ M N P \equiv Δ M Q P) \\ {(10)}^{2} & = x^{2} + {(8)}^{2} (Pythagoras) \\ 100 & = x^{2} + 64 \\ 100 - 64 & = x^{2} \\ \pm \sqrt{36} & = x \\ 6 & = x \end{aligned}

$\begin{aligned} \text{In } & \Delta MQP: \quad \quad\\ Q P & = 10 \quad\quad (\Delta MNP \equiv \Delta MQP) \\ \left( 10 \right)^{2} & = x^{2} + \left( 8 \right)^{2} \quad (\text{Pythagoras})\\ 100 & = x^{2} + 64 \\ 100 - 64 & = x^{2} \\ \pm \sqrt{ 36 } &= x \\ 6 & = x \end{aligned}$

The length of $MQ$ is $\text{6}\ \text{units}$ .

STEP: Find the length of

Q N

$QN$
[−1 point ⇒ 0 / 4 points left]

For the second question, we need to find the length of segment $QN$ . This segment is twice as long as $x$ , which we calculated above. So $QN = 2 \left( 6 \right) = \text{12}\ \text{units}$ .

The correct answers are:

The length of side $x$ is $\text{6}\ \text{units}$ .
The length of $QN$ is $\text{12}\ \text{units}$ .

Submit your answer as: and

ID is: 1858 Seed is: 8337

Congruent triangles

In the diagram below, $\Delta ABC \equiv \Delta ADC$ . Also, $AC \perp DB$ while $BC = 30$ and $AC = 24$ .

Calculate the value of $x$ .
Determine the length of $DB$ .

Answer:

$x =$ $\text{units}$
$DB =$ $\text{units}$

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

The triangles are congruent. That means they are identical. Start by using this fact to label all the sides which are not already labelled.

STEP: Label everything you can in the figure
[−1 point ⇒ 3 / 4 points left]

The first question asks us to determine the length of the side of the diagram labelled $x$ . The $x$ is attached to side $AD$ .

Based on the labels in the diagram, we do not know enough to calculate $x$ immediately. However, the question states that the two triangles are congruent. That means they are identical. We can use that information to label all the sides which are not already labelled. The question also states that $AC \perp DB$ , which means that there are two right angles. We should label those right angles too.

Now we can see two right-angled triangles. And we can see which sides are equal to each other. Both of the triangles include a side labelled $x$ .

STEP: Use the theorem of Pythagoras to find the value of

x

$x$
[−2 points ⇒ 1 / 4 points left]

Now calculate $x$ using the theorem of Pythagoras. We will focus on triangle $ADC$ , but you can use either one, because they are congruent.

\begin{aligned} In & Δ A D C : \\ D C & = 30 (Δ A B C \equiv Δ A D C) \\ {(30)}^{2} & = x^{2} + {(24)}^{2} (Pythagoras) \\ 900 & = x^{2} + 576 \\ 900 - 576 & = x^{2} \\ \pm \sqrt{324} & = x \\ 18 & = x \end{aligned}

$\begin{aligned} \text{In } & \Delta ADC: \quad \quad\\ D C & = 30 \quad\quad (\Delta ABC \equiv \Delta ADC) \\ \left( 30 \right)^{2} & = x^{2} + \left( 24 \right)^{2} \quad (\text{Pythagoras})\\ 900 & = x^{2} + 576 \\ 900 - 576 & = x^{2} \\ \pm \sqrt{ 324 } &= x \\ 18 & = x \end{aligned}$

The length of $AD$ is $\text{18}\ \text{units}$ .

STEP: Find the length of

D B

$DB$
[−1 point ⇒ 0 / 4 points left]

For the second question, we need to find the length of segment $DB$ . This segment is twice as long as $x$ , which we calculated above. So $DB = 2 \left( 18 \right) = \text{36}\ \text{units}$ .

The correct answers are:

The length of side $x$ is $\text{18}\ \text{units}$ .
The length of $DB$ is $\text{36}\ \text{units}$ .

Submit your answer as: and

ID is: 1858 Seed is: 873

Congruent triangles

In the diagram below, $\Delta ABC \equiv \Delta ADC$ . Also, $AC \perp DB$ while $BC = 10$ and $AC = 8$ .

Calculate the value of $x$ .
Determine the length of $DB$ .

Answer:

$x =$ $\text{units}$
$DB =$ $\text{units}$

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

The triangles are congruent. That means they are identical. Start by using this fact to label all the sides which are not already labelled.

STEP: Label everything you can in the figure
[−1 point ⇒ 3 / 4 points left]

The first question asks us to determine the length of the side of the diagram labelled $x$ . The $x$ is attached to side $AD$ .

Based on the labels in the diagram, we do not know enough to calculate $x$ immediately. However, the question states that the two triangles are congruent. That means they are identical. We can use that information to label all the sides which are not already labelled. The question also states that $AC \perp DB$ , which means that there are two right angles. We should label those right angles too.

Now we can see two right-angled triangles. And we can see which sides are equal to each other. Both of the triangles include a side labelled $x$ .

STEP: Use the theorem of Pythagoras to find the value of

x

$x$
[−2 points ⇒ 1 / 4 points left]

Now calculate $x$ using the theorem of Pythagoras. We will focus on triangle $ADC$ , but you can use either one, because they are congruent.

\begin{aligned} In & Δ A D C : \\ D C & = 10 (Δ A B C \equiv Δ A D C) \\ {(10)}^{2} & = x^{2} + {(8)}^{2} (Pythagoras) \\ 100 & = x^{2} + 64 \\ 100 - 64 & = x^{2} \\ \pm \sqrt{36} & = x \\ 6 & = x \end{aligned}

$\begin{aligned} \text{In } & \Delta ADC: \quad \quad\\ D C & = 10 \quad\quad (\Delta ABC \equiv \Delta ADC) \\ \left( 10 \right)^{2} & = x^{2} + \left( 8 \right)^{2} \quad (\text{Pythagoras})\\ 100 & = x^{2} + 64 \\ 100 - 64 & = x^{2} \\ \pm \sqrt{ 36 } &= x \\ 6 & = x \end{aligned}$

The length of $AD$ is $\text{6}\ \text{units}$ .

STEP: Find the length of

D B

$DB$
[−1 point ⇒ 0 / 4 points left]

For the second question, we need to find the length of segment $DB$ . This segment is twice as long as $x$ , which we calculated above. So $DB = 2 \left( 6 \right) = \text{12}\ \text{units}$ .

The correct answers are:

The length of side $x$ is $\text{6}\ \text{units}$ .
The length of $DB$ is $\text{12}\ \text{units}$ .

Submit your answer as: and

ID is: 4182 Seed is: 6904

Definitions of quadrilaterals

Select the most correct definition for a parallelogram.
Answer: A parallelogram is a:
tilted rectangle.
quadrilateral with at least one pair of opposite sides parallel.
quadrilateral with two pairs of adjacent sides equal.
quadrilateral with two pairs of opposite sides parallel.
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to choose the definition that lists all the properties which make the shape what it is. None of the essential properties should be missing from your definition. You also must not include any unnecessary properties.

STEP: Choose the definition with the exact properties
[−2 points ⇒ 0 / 2 points left]

We use the properties of shapes to define them. We need to choose the definition that lists all the properties which make the shape what it is. None of the essential properties should be missing from the definition. Also, the definition must not include any unnecessary properties.
- ✘ "Tilted rectangle" is not a useful way to think about a parallelogram. It suggests that a parallelogram is a type of rectangle, but this is not true.
- ✘ A quadrilateral with at least one pair of opposite sides parallel is a trapezium.
- ✘ A quadrilateral with two pairs of adjacent sides equal is a kite.
- ✔ A parallelogram is a quadrilateral with two pairs of opposite sides parallel.
This is a diagram of a parallelogram with the definition properties drawn in:

Submit your answer as:
quadrilateral with two pairs of opposite sides parallel.
Complete the statement below:

Answer:
A has all of the same properties as a parallelogram (plus some more), so it is a special type of parallelogram.

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Look at the definition for a parallelogram, and ask yourself which quadrilateral on the list has these same properties. It can have extra properties, but it cannot have less.

STEP: Choose the quadrilateral that has at least the same properties as a parallelogram
[−1 point ⇒ 0 / 1 points left]

A square has two pairs of opposite sides parallel. This means that a square meets the definition of a parallelogram, so we say that it is a special type of parallelogram.

NOTE: It does not matter that a square has extra properties: what matters is that it has enough properties to be called a parallelogram.

Submit your answer as:

ID is: 4182 Seed is: 4030

Definitions of quadrilaterals

Select the most correct definition for a kite.
Answer: A kite is a:
quadrilateral with two pairs of adjacent sides equal.
quadrilateral with two pairs of opposite sides parallel.
quadrilateral in the shape of a diamond.
quadrilateral with two pairs of opposite sides parallel and all corners $\text{90}\text{°}$ .
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to choose the definition that lists all the properties which make the shape what it is. None of the essential properties should be missing from your definition. You also must not include any unnecessary properties.

STEP: Choose the definition with the exact properties
[−2 points ⇒ 0 / 2 points left]

We use the properties of shapes to define them. We need to choose the definition that lists all the properties which make the shape what it is. None of the essential properties should be missing from the definition. Also, the definition must not include any unnecessary properties.
- ✔ A kite is a quadrilateral with two pairs of adjacent sides equal.
- ✘ A quadrilateral with two pairs of opposite sides parallel is a parallelogram.
- ✘ We do not refer to any quadrilaterals as "diamonds". This is an informal name. As we progress further in mathematics, we need to start to use the correct terms.
- ✘ A quadrilateral with two pairs of opposite sides parallel and all corners $\text{90}\text{°}$ is a rectangle.
This is a diagram of a kite with the definition properties drawn in:

Submit your answer as:
quadrilateral with two pairs of adjacent sides equal.
Complete the statement below:

Answer:
A has all of the same properties as a kite (plus some more), so it is a special type of kite.

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Look at the definition for a kite, and ask yourself which quadrilateral on the list has these same properties. It can have extra properties, but it cannot have less.

STEP: Choose the quadrilateral that has at least the same properties as a kite
[−1 point ⇒ 0 / 1 points left]

A square has two pairs of adjacent sides equal. This means that a square meets the definition of a kite, so we say that it is a special type of kite.

NOTE: It does not matter that a square has extra properties: what matters is that it has enough properties to be called a kite.

Submit your answer as:

ID is: 4182 Seed is: 8567

Definitions of quadrilaterals

Select the most correct definition for a rhombus.
Answer: A rhombus is a:
quadrilateral with two pairs of opposite sides parallel and adjacent sides equal.
tilted square.
quadrilateral with two pairs of opposite sides parallel, adjacent sides equal, and all corners $\text{90}\text{°}$ .
quadrilateral with two pairs of opposite sides parallel and all corners $\text{90}\text{°}$ .
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to choose the definition that lists all the properties which make the shape what it is. None of the essential properties should be missing from your definition. You also must not include any unnecessary properties.

STEP: Choose the definition with the exact properties
[−2 points ⇒ 0 / 2 points left]

We use the properties of shapes to define them. We need to choose the definition that lists all the properties which make the shape what it is. None of the essential properties should be missing from the definition. Also, the definition must not include any unnecessary properties.
- ✔ A rhombus is a quadrilateral with two pairs of opposite sides parallel and adjacent sides equal.
- ✘ "Tilted square" is not a useful way to think about a rhombus. It suggests that a rhombus is a type of square, but this is not true.
- ✘ A quadrilateral with two pairs of opposite sides parallel, adjacent sides equal, and all corners $\text{90}\text{°}$ is a square.
- ✘ A quadrilateral with two pairs of opposite sides parallel and all corners $\text{90}\text{°}$ is a rectangle.
This is a diagram of a rhombus with the definition properties drawn in:

Submit your answer as:
quadrilateral with two pairs of opposite sides parallel and adjacent sides equal.
Complete the statement below:

Answer:
A has all of the same properties as a rhombus (plus some more), so it is a special type of rhombus.

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Look at the definition for a rhombus, and ask yourself which quadrilateral on the list has these same properties. It can have extra properties, but it cannot have less.

STEP: Choose the quadrilateral that has at least the same properties as a rhombus
[−1 point ⇒ 0 / 1 points left]

A square has two pairs of opposite sides parallel and adjacent sides equal. This means that a square meets the definition of a rhombus, so we say that it is a special type of rhombus.

NOTE: It does not matter that a square has extra properties: what matters is that it has enough properties to be called a rhombus.

Submit your answer as:

ID is: 1853 Seed is: 2170

Opposites of a parallelogram

Answer the following questions about the parallelogram $XWVU$ . The parallelogram has the following sides and angles:

$\hat X = 63°$ , $\hat W = x$ , $\hat V = y$ , and $\hat U = 117°$ . The sides are ${XW} = w$ , ${WV} = 4$ , ${VU} = 7$ , and ${XU} = 4$ .

Answer:

What is the size of $x$ ? °
What is the size of $y$ ? °
What is the length of side $w$ ? $\text{units}$

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

You do not need to calculate anything. You can read all three answers from the diagram. You need to use the properties of sides and angles for parallelograms.

STEP: Find the unknown angles and the length of the missing side
[−3 points ⇒ 0 / 3 points left]

We can answer this question based on the properties of parallelograms.

There are two properties of parallelograms which are useful in this question:

Opposite angles in a parallelogram are equal
Opposite sides of parallelograms have equal lengths

Therefore:

${\color{red}{x = 117\text{°}}}$
${\color{blue}{y = 63\text{°}}}$
${\color{ForestGreen}{w = 7 \; \text{units}}}$

Submit your answer as: andand

ID is: 1853 Seed is: 1483

Opposites of a parallelogram

Answer the following questions about the parallelogram $XWVU$ . The parallelogram has the following sides and angles:

$\hat X = y$ , $\hat W = x$ , $\hat V = 39°$ , and $\hat U = 141°$ . The sides are ${XW} = 6$ , ${WV} = w$ , ${VU} = 6$ , and ${XU} = 6$ .

Answer:

What is the size of $x$ ? °
What is the size of $y$ ? °
What is the length of side $w$ ? $\text{units}$

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

You do not need to calculate anything. You can read all three answers from the diagram. You need to use the properties of sides and angles for parallelograms.

STEP: Find the unknown angles and the length of the missing side
[−3 points ⇒ 0 / 3 points left]

We can answer this question based on the properties of parallelograms.

There are two properties of parallelograms which are useful in this question:

Opposite angles in a parallelogram are equal
Opposite sides of parallelograms have equal lengths

Therefore:

${\color{red}{x = 141\text{°}}}$
${\color{blue}{y = 39\text{°}}}$
${\color{ForestGreen}{w = 6 \; \text{units}}}$

Submit your answer as: andand

ID is: 1853 Seed is: 6630

Opposites of a parallelogram

Answer the following questions about the parallelogram $XWVU$ . The parallelogram has the following sides and angles:

$\hat X = 74°$ , $\hat W = y$ , $\hat V = x$ , and $\hat U = 106°$ . The sides are ${XW} = 5$ , ${WV} = 7$ , ${VU} = 5$ , and ${XU} = w$ .

Answer:

What is the size of $x$ ? °
What is the size of $y$ ? °
What is the length of side $w$ ? $\text{units}$

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

You do not need to calculate anything. You can read all three answers from the diagram. You need to use the properties of sides and angles for parallelograms.

STEP: Find the unknown angles and the length of the missing side
[−3 points ⇒ 0 / 3 points left]

We can answer this question based on the properties of parallelograms.

There are two properties of parallelograms which are useful in this question:

Opposite angles in a parallelogram are equal
Opposite sides of parallelograms have equal lengths

Therefore:

${\color{blue}{x = 74\text{°}}}$
${\color{red}{y = 106\text{°}}}$
${\color{ForestGreen}{w = 7 \; \text{units}}}$

Submit your answer as: andand

ID is: 4216 Seed is: 7555

Identifying quadrilaterals

Which of the following are kites?

A	B

C	D

Answer: The kites are .

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Which of these quadrilaterals have the properties which meet the definition of a kites? They could have extra properties as well - that does not mean they are not kites.

STEP: Identify the quadrilaterals that meet the definition of a kite
[−2 points ⇒ 0 / 2 points left]

A kite is a quadrilateral with two pairs of equal adjacent sides.

We can easily recognise Quadrilateral D as a kite. However, this is not the only type of kite in the options.

Quadrilaterals A and C each have two pairs of equal adjacent sides. So, A and C are also types of kites. This is true even though the exact name for A is a square, and the exact name for C is a rhombus (because they have a few extra properties too).

Submit your answer as:

ID is: 4216 Seed is: 4032

Identifying quadrilaterals

Which of the following are parallelograms?

A	B

C	D

Answer: The parallelograms are .

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Which of these quadrilaterals have the properties which meet the definition of a parallelograms? They could have extra properties as well - that does not mean they are not parallelograms.

STEP: Identify the quadrilaterals that meet the definition of a parallelogram
[−2 points ⇒ 0 / 2 points left]

A parallelogram is a quadrilateral with two pairs of opposite sides that are parallel.

We can easily recognise Quadrilateral B as a parallelogram. However, this is not the only type of parallelogram in the options.

Quadrilaterals C and D each have two pairs of opposite sides that are parallel. So, C and D are also types of parallelograms. This is true even though the exact name for C is a rectangle, and the exact name for D is a square (because they have a few extra properties too).

Submit your answer as:

ID is: 4216 Seed is: 7590

Identifying quadrilaterals

Which of the following are parallelograms?

A	B

C	D

Answer: The parallelograms are .

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Which of these quadrilaterals have the properties which meet the definition of a parallelograms? They could have extra properties as well - that does not mean they are not parallelograms.

STEP: Identify the quadrilaterals that meet the definition of a parallelogram
[−2 points ⇒ 0 / 2 points left]

A parallelogram is a quadrilateral with two pairs of opposite sides that are parallel.

We can easily recognise Quadrilateral A as a parallelogram. However, this is not the only type of parallelogram in the options.

Quadrilaterals B and C each have two pairs of opposite sides that are parallel. So, B and C are also types of parallelograms. This is true even though the exact name for B is a rhombus, and the exact name for C is a square (because they have a few extra properties too).

Submit your answer as:

ID is: 1846 Seed is: 4706

Angles on a straight line

Line $\overleftrightarrow{BC}$ represents angles on one side of a straight line. $\angle a$ = $58°$ , $\angle b$ = $x$ and $\angle c$ = $58°$ .

Answer the following questions about the diagram:

What is the value of $\angle x$ ?
What type of angle is represented by $x$ ?

Answer:

$x =$ ${°}$
The angle is .

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

It can be shown that the measurement of a straight angle is $180{°}$ or $\pi$ radians.

STEP: Use the fact that angles on a straight line add to

180 °

$180°$
[−2 points ⇒ 0 / 2 points left]

Flat or straight angles are formed when the legs are pointing in exactly opposite directions. The two legs then form a single straight line through the vertex of the angle. Angles in a straight line add up to $180{°}$ .

\begin{aligned} ∠ a + ∠ b + ∠ c & = 180 ° \\ 58 ° + x + 58 ° & = 180 ° \\ ∴ x & = 64 ° \end{aligned}

$\begin{aligned} \angle a + \angle b + \angle c & = 180 {°} \\ 58° + x + 58° & = 180 {°} \\ \therefore x & = 64° \end{aligned}$

Submit your answer as: and

ID is: 1846 Seed is: 6883

Angles on a straight line

Line $\overleftrightarrow{YZ}$ represents angles on one side of a straight line. $\angle a$ = $46°$ , $\angle b$ = $54°$ and $\angle c$ = $x$ .

Answer the following questions about the diagram:

What is the value of $\angle x$ ?
What type of angle is represented by $x$ ?

Answer:

$x =$ ${°}$
The angle is .

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

It can be shown that the measurement of a straight angle is $180{°}$ or $\pi$ radians.

STEP: Use the fact that angles on a straight line add to

180 °

$180°$
[−2 points ⇒ 0 / 2 points left]

\begin{aligned} ∠ a + ∠ b + ∠ c & = 180 ° \\ 46 ° + 54 ° + x & = 180 ° \\ ∴ x & = 80 ° \end{aligned}

$\begin{aligned} \angle a + \angle b + \angle c & = 180 {°} \\ 46° + 54° + x & = 180 {°} \\ \therefore x & = 80° \end{aligned}$

Submit your answer as: and

ID is: 1846 Seed is: 8263

Angles on a straight line

Line $\overleftrightarrow{BC}$ represents angles on one side of a straight line. $\angle a$ = $68°$ , $\angle b$ = $70°$ and $\angle c$ = $x$ .

Answer the following questions about the diagram:

What is the value of $\angle x$ ?
What type of angle is represented by $x$ ?

Answer:

$x =$ ${°}$
The angle is .

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

It can be shown that the measurement of a straight angle is $180{°}$ or $\pi$ radians.

STEP: Use the fact that angles on a straight line add to

180 °

$180°$
[−2 points ⇒ 0 / 2 points left]

\begin{aligned} ∠ a + ∠ b + ∠ c & = 180 ° \\ 68 ° + 70 ° + x & = 180 ° \\ ∴ x & = 42 ° \end{aligned}

$\begin{aligned} \angle a + \angle b + \angle c & = 180 {°} \\ 68° + 70° + x & = 180 {°} \\ \therefore x & = 42° \end{aligned}$

Submit your answer as: and

ID is: 1864 Seed is: 2508

Angles on a straight line

Examine the diagram below, and answer the questions that follow.

Calculate $x$ .
What type of angle is represented with $x$ ?

Answer:

$x =$ $°$
The angle is: .

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

The angles must have a sum of $180°$ . Write an equation based on this information and solve for $x$ .

STEP: Write an equation to summarise the angles
[−1 point ⇒ 1 / 2 points left]

The three angles must have a sum of $180°$ because $BC$ is a straight line. We can write an equation based on this information:

\begin{aligned} x + 108 ° + 40 ° & = 180 ° (∠ s on a str line) \end{aligned}

$\begin{aligned} x + 108° + 40° & = 180° \quad (\angle \text{s on a str line}) \end{aligned}$

STEP: Solve the equation and deterimine the type of the angle
[−1 point ⇒ 0 / 2 points left]

Now we can solve the equation. Once we have the answer we can determine the type of angle.

\begin{aligned} x + 108 ° + 40 ° & = 180 ° \\ ∴ x & = 32 ° \end{aligned}

$\begin{aligned} x + 108° + 40° & = 180 {°} \\ \therefore x & = 32° \end{aligned}$

The complete diagram, with all three angles known, is:

The missing angle is $32°$ , which is acute.

Submit your answer as: and

ID is: 1864 Seed is: 2102

Angles on a straight line

Examine the diagram below, and answer the questions that follow.

Calculate $x$ .
What type of angle is represented with $x$ ?

Answer:

$x =$ $°$
The angle is: .

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

The angles must have a sum of $180°$ . Write an equation based on this information and solve for $x$ .

STEP: Write an equation to summarise the angles
[−1 point ⇒ 1 / 2 points left]

The three angles must have a sum of $180°$ because $YZ$ is a straight line. We can write an equation based on this information:

\begin{aligned} 50 ° + x + 50 ° & = 180 ° (∠ s on a str line) \end{aligned}

$\begin{aligned} 50° + x + 50° & = 180° \quad (\angle \text{s on a str line}) \end{aligned}$

STEP: Solve the equation and deterimine the type of the angle
[−1 point ⇒ 0 / 2 points left]

Now we can solve the equation. Once we have the answer we can determine the type of angle.

\begin{aligned} 50 ° + x + 50 ° & = 180 ° \\ ∴ x & = 80 ° \end{aligned}

$\begin{aligned} 50° + x + 50° & = 180 {°} \\ \therefore x & = 80° \end{aligned}$

The complete diagram, with all three angles known, is:

The missing angle is $80°$ , which is acute.

Submit your answer as: and

ID is: 1864 Seed is: 6528

Angles on a straight line

Examine the diagram below, and answer the questions that follow.

Calculate $x$ .
What type of angle is represented with $x$ ?

Answer:

$x =$ $°$
The angle is: .

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

The angles must have a sum of $180°$ . Write an equation based on this information and solve for $x$ .

STEP: Write an equation to summarise the angles
[−1 point ⇒ 1 / 2 points left]

The three angles must have a sum of $180°$ because $ST$ is a straight line. We can write an equation based on this information:

\begin{aligned} 57 ° + x + 60 ° & = 180 ° (∠ s on a str line) \end{aligned}

$\begin{aligned} 57° + x + 60° & = 180° \quad (\angle \text{s on a str line}) \end{aligned}$

STEP: Solve the equation and deterimine the type of the angle
[−1 point ⇒ 0 / 2 points left]

Now we can solve the equation. Once we have the answer we can determine the type of angle.

\begin{aligned} 57 ° + x + 60 ° & = 180 ° \\ ∴ x & = 63 ° \end{aligned}

$\begin{aligned} 57° + x + 60° & = 180 {°} \\ \therefore x & = 63° \end{aligned}$

The complete diagram, with all three angles known, is:

The missing angle is $63°$ , which is acute.

Submit your answer as: and

ID is: 1672 Seed is: 3157

Prove the quadrilateral is a parallelogram

The figure below shows quadrilateral $XWVU$ , which is a parallelogram. The $\text{4}$ dashed red lines bisect each vertex (angle) of the parallelogram (angles $X, W, V$ and $U$ ). So $X_{1} = X_{2}$ , $W_{1} = W_{2}$ , and so on. These dashed lines form quadrilateral $EFGH$ inside of parallelogram $XWVU$ .

Note the diagram is drawn to scale.

The table below shows a proof that quadrilaterial $EFGH$ is a parallelogram. But $\text{4}$ steps and $\text{4}$ reasons are missing from the proof. Complete the proof by choosing the correct steps and reasons from each of the drop down boxes.

Answer:

STEPS	REASONS
$\angle X = \angle V$	Opposite angles in a parallelogram are equal

$\angle W = \angle U$	Opposite angles in a parallelogram are equal
$\angle W_{2} = \angle U_{1}$ and $\angle W_{1} = \angle U_{2}$	Angles created by bisectors of equal angles are equal
$XU = WV$	Opposite sides of a parallelogram are equal

In quadrilateral $EFGH: \angle E_{2} = \angle G_{2}$	Congruent triangles have equal angles $(\Delta XUE \equiv \Delta VGW)$


$\angle H_{1} = \angle F_{1}$	Congruent triangles have equal angles $(\Delta XHW \equiv \Delta VFU)$
$\angle H_{1} = \angle H_{2}$ and $\angle F_{1} = \angle F_{2}$	Vertically opposite angles are equal
$EFGH$ is a parallelogram	Opposite angles in a parallelogram are equal

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 8 / 8 points left]

You should start at the top of the proof and work your way down. Try to understand each step and each reason, because you will need to know how the proof is developing from the top to the bottom.

STEP: Work down the proof and select the steps and reasons along the way
[−8 points ⇒ 0 / 8 points left]

The proof is about the quadrilateral in the middle of the figure, $EFGH$ . The steps of the proof need to show that this quadrilateral is a parallelogram. We need to complete the proof by selecting the correct steps and reasons.

The question states that quadrilateral $XWVU$ is a parallelogram. We know lots of facts about parallelograms. The question also states that the red dashed lines bisect the angles at each vertex. (This means the line splits the angle into two equal parts.) It is a good idea to mark everything we know on the diagram itself. The diagram below shows a number of new labels for equal angles and equal sides.

The strategy used in this proof can be summarised as follows: we want to prove that $EFGH$ is a parallelogram. We can do that if we can show that the opposite angles inside $EFGH$ are equal. So the proof shows that there are congruent triangles in the diagram. That leads to congruent angles. And from those congruent angles we can conclude that $EFGH$ has two pairs of opposite angles which are equal.

NOTE: There are other ways to prove that quadrilateral

E F G H

$EFGH$ is a parallelogram: this is not the only way to do it. On a test or exam, you might use a different set of steps and reasons. But for this question, you must find the steps and reasons which fit within the proof as it is presented.

The table below shows the correct answer choices (in the green blocks).

STEPS	REASONS
$\angle X = \angle V$	Opposite angles in a parallelogram are equal
$\angle X_{2} = \angle V_{1}$ and $\angle X_{1} = \angle V_{2}$	Angles created by bisectors of equal angles are equal
$\angle W = \angle U$	Opposite angles in a parallelogram are equal
$\angle W_{2} = \angle U_{1}$ and $\angle W_{1} = \angle U_{2}$	Angles created by bisectors of equal angles are equal
$XU = WV$	Opposite sides of a parallelogram are equal
$\Delta XUE \equiv \Delta VGW$	Congruent triangles (Angle-Side-Angle)
In quadrilateral $EFGH: \angle E_{2} = \angle G_{2}$	Congruent triangles have equal angles $(\Delta XUE \equiv \Delta VGW)$
$XW = UV$	Opposite sides of a parallelogram are equal
$\Delta XHW \equiv \Delta VFU$	Congruent triangles (Angle-Side-Angle)
$\angle H_{1} = \angle F_{1}$	Congruent triangles have equal angles $(\Delta XHW \equiv \Delta VFU)$
$\angle H_{1} = \angle H_{2}$ and $\angle F_{1} = \angle F_{2}$	Vertically opposite angles are equal
$EFGH$ is a parallelogram	Opposite angles in a parallelogram are equal

Submit your answer as: andandandandandandand

ID is: 1672 Seed is: 7478

Prove the quadrilateral is a parallelogram

The figure below shows quadrilateral $ABCD$ , which is a parallelogram. The $\text{4}$ dashed red lines bisect each vertex (angle) of the parallelogram (angles $A, B, C$ and $D$ ). So $A_{1} = A_{2}$ , $B_{1} = B_{2}$ , and so on. These dashed lines form quadrilateral $MNOP$ inside of parallelogram $ABCD$ .

Note the diagram is drawn to scale.

The table below shows a proof that quadrilaterial $MNOP$ is a parallelogram. But $\text{4}$ steps and $\text{4}$ reasons are missing from the proof. Complete the proof by choosing the correct steps and reasons from each of the drop down boxes.

Answer:

STEPS	REASONS
$\angle A = \angle C$
$\angle A_{2} = \angle C_{1}$ and $\angle A_{1} = \angle C_{2}$
$\angle B = \angle D$	Opposite angles in a parallelogram are equal
	Angles created by bisectors of equal angles are equal
$AD = BC$	Opposite sides of a parallelogram are equal
	Congruent triangles (Angle-Side-Angle)
	Congruent triangles have equal angles $(\Delta ADM \equiv \Delta COB)$
$AB = DC$	Opposite sides of a parallelogram are equal
$\Delta APB \equiv \Delta CND$
	Congruent triangles have equal angles $(\Delta APB \equiv \Delta CND)$
$\angle P_{1} = \angle P_{2}$ and $\angle N_{1} = \angle N_{2}$
$MNOP$ is a parallelogram	Opposite angles in a parallelogram are equal

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 8 / 8 points left]

You should start at the top of the proof and work your way down. Try to understand each step and each reason, because you will need to know how the proof is developing from the top to the bottom.

STEP: Work down the proof and select the steps and reasons along the way
[−8 points ⇒ 0 / 8 points left]

The proof is about the quadrilateral in the middle of the figure, $MNOP$ . The steps of the proof need to show that this quadrilateral is a parallelogram. We need to complete the proof by selecting the correct steps and reasons.

The question states that quadrilateral $ABCD$ is a parallelogram. We know lots of facts about parallelograms. The question also states that the red dashed lines bisect the angles at each vertex. (This means the line splits the angle into two equal parts.) It is a good idea to mark everything we know on the diagram itself. The diagram below shows a number of new labels for equal angles and equal sides.

The strategy used in this proof can be summarised as follows: we want to prove that $MNOP$ is a parallelogram. We can do that if we can show that the opposite angles inside $MNOP$ are equal. So the proof shows that there are congruent triangles in the diagram. That leads to congruent angles. And from those congruent angles we can conclude that $MNOP$ has two pairs of opposite angles which are equal.

NOTE: There are other ways to prove that quadrilateral

M N O P

$MNOP$ is a parallelogram: this is not the only way to do it. On a test or exam, you might use a different set of steps and reasons. But for this question, you must find the steps and reasons which fit within the proof as it is presented.

The table below shows the correct answer choices (in the green blocks).

STEPS	REASONS
$\angle A = \angle C$	Opposite angles in a parallelogram are equal
$\angle A_{2} = \angle C_{1}$ and $\angle A_{1} = \angle C_{2}$	Angles created by bisectors of equal angles are equal
$\angle B = \angle D$	Opposite angles in a parallelogram are equal
$\angle B_{2} = \angle D_{1}$ and $\angle B_{1} = \angle D_{2}$	Angles created by bisectors of equal angles are equal
$AD = BC$	Opposite sides of a parallelogram are equal
$\Delta ADM \equiv \Delta COB$	Congruent triangles (Angle-Side-Angle)
In quadrilateral $MNOP: \angle M_{2} = \angle O_{2}$	Congruent triangles have equal angles $(\Delta ADM \equiv \Delta COB)$
$AB = DC$	Opposite sides of a parallelogram are equal
$\Delta APB \equiv \Delta CND$	Congruent triangles (Angle-Side-Angle)
$\angle P_{1} = \angle N_{1}$	Congruent triangles have equal angles $(\Delta APB \equiv \Delta CND)$
$\angle P_{1} = \angle P_{2}$ and $\angle N_{1} = \angle N_{2}$	Vertically opposite angles are equal
$MNOP$ is a parallelogram	Opposite angles in a parallelogram are equal

Submit your answer as: andandandandandandand

ID is: 1672 Seed is: 8308

Prove the quadrilateral is a parallelogram

The figure below shows quadrilateral $QRST$ , which is a parallelogram. The $\text{4}$ dashed red lines bisect each vertex (angle) of the parallelogram (angles $Q, R, S$ and $T$ ). So $Q_{1} = Q_{2}$ , $R_{1} = R_{2}$ , and so on. These dashed lines form quadrilateral $JKLM$ inside of parallelogram $QRST$ .

Note the diagram is drawn to scale.

The table below shows a proof that quadrilaterial $JKLM$ is a parallelogram. But $\text{4}$ steps and $\text{4}$ reasons are missing from the proof. Complete the proof by choosing the correct steps and reasons from each of the drop down boxes.

Answer:

STEPS	REASONS

$\angle Q_{2} = \angle S_{1}$ and $\angle Q_{1} = \angle S_{2}$	Angles created by bisectors of equal angles are equal
$\angle R = \angle T$	Opposite angles in a parallelogram are equal
	Angles created by bisectors of equal angles are equal
$QT = RS$	Opposite sides of a parallelogram are equal

In quadrilateral $JKLM: \angle J_{2} = \angle L_{2}$	Congruent triangles have equal angles $(\Delta QTJ \equiv \Delta SLR)$
$QR = TS$	Opposite sides of a parallelogram are equal
$\Delta QMR \equiv \Delta SKT$	Congruent triangles (Angle-Side-Angle)
	Congruent triangles have equal angles $(\Delta QMR \equiv \Delta SKT)$
$\angle M_{1} = \angle M_{2}$ and $\angle K_{1} = \angle K_{2}$
$JKLM$ is a parallelogram

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 8 / 8 points left]

You should start at the top of the proof and work your way down. Try to understand each step and each reason, because you will need to know how the proof is developing from the top to the bottom.

STEP: Work down the proof and select the steps and reasons along the way
[−8 points ⇒ 0 / 8 points left]

The proof is about the quadrilateral in the middle of the figure, $JKLM$ . The steps of the proof need to show that this quadrilateral is a parallelogram. We need to complete the proof by selecting the correct steps and reasons.

The question states that quadrilateral $QRST$ is a parallelogram. We know lots of facts about parallelograms. The question also states that the red dashed lines bisect the angles at each vertex. (This means the line splits the angle into two equal parts.) It is a good idea to mark everything we know on the diagram itself. The diagram below shows a number of new labels for equal angles and equal sides.

The strategy used in this proof can be summarised as follows: we want to prove that $JKLM$ is a parallelogram. We can do that if we can show that the opposite angles inside $JKLM$ are equal. So the proof shows that there are congruent triangles in the diagram. That leads to congruent angles. And from those congruent angles we can conclude that $JKLM$ has two pairs of opposite angles which are equal.

NOTE: There are other ways to prove that quadrilateral

J K L M

$JKLM$ is a parallelogram: this is not the only way to do it. On a test or exam, you might use a different set of steps and reasons. But for this question, you must find the steps and reasons which fit within the proof as it is presented.

The table below shows the correct answer choices (in the green blocks).

STEPS	REASONS
$\angle Q = \angle S$	Opposite angles in a parallelogram are equal
$\angle Q_{2} = \angle S_{1}$ and $\angle Q_{1} = \angle S_{2}$	Angles created by bisectors of equal angles are equal
$\angle R = \angle T$	Opposite angles in a parallelogram are equal
$\angle R_{2} = \angle T_{1}$ and $\angle R_{1} = \angle T_{2}$	Angles created by bisectors of equal angles are equal
$QT = RS$	Opposite sides of a parallelogram are equal
$\Delta QTJ \equiv \Delta SLR$	Congruent triangles (Angle-Side-Angle)
In quadrilateral $JKLM: \angle J_{2} = \angle L_{2}$	Congruent triangles have equal angles $(\Delta QTJ \equiv \Delta SLR)$
$QR = TS$	Opposite sides of a parallelogram are equal
$\Delta QMR \equiv \Delta SKT$	Congruent triangles (Angle-Side-Angle)
$\angle M_{1} = \angle K_{1}$	Congruent triangles have equal angles $(\Delta QMR \equiv \Delta SKT)$
$\angle M_{1} = \angle M_{2}$ and $\angle K_{1} = \angle K_{2}$	Vertically opposite angles are equal
$JKLM$ is a parallelogram	Opposite angles in a parallelogram are equal

Submit your answer as: andandandandandandand

ID is: 1866 Seed is: 4127

Angles in a full turn

In the diagram below, line $YZ$ is a straight line with angles labelled above and below it.

Answer the following questions about the diagram:

Answer:

$x =$ $°$
$y =$ $°$
$y$ is .

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Angles on a straight line add up to $180°$ .

STEP: <no title>
[−3 points ⇒ 0 / 3 points left]

Line $YZ$ is a straight line, and angles on a straight line up to $180°$ . This means that all of the angles above $YZ$ add up to $180°$ :

\begin{aligned} 10 ° + x + 98 ° & = 180 ° (∠ s on a str line) \\ ∴ x & = 72 ° \end{aligned}

$\begin{aligned} 10° + x + 98° & = 180{°} \quad (\angle \text{s on a str line}) \\ \therefore x & = 72° \end{aligned}$

In the same way, all of the angles below $YZ$ add up to $180°$ :

\begin{aligned} 34 ° + 58 ° + y + 51 ° & = 180 ° (∠ s on a str line) \\ ∴ y & = 37 ° \end{aligned}

$\begin{aligned} 34° + 58° + y + 51° & = 180{°} \quad (\angle \text{s on a str line}) \\ \therefore y & = 37° \end{aligned}$

Here is the completed diagram, with all of the angle values labelled:

The angles in a full turn (revolution) must add up to $360°$ . This is a good way to check our answers:

\begin{aligned} 10 ° + 72 ° + 98 ° + 34 ° + 58 ° + 37 ° + 51 ° & = 360 ° \end{aligned}

$\begin{aligned} 10° + 72° + 98° + 34° + 58° + 37° + 51° & = 360{°} \end{aligned}$

$y = 37°$

$\therefore y$ is an acute angle (less than 90 $°$ ).

Submit your answer as: andand

ID is: 1866 Seed is: 2125

Angles in a full turn

In the diagram below, line $YZ$ is a straight line with angles labelled above and below it.

Answer the following questions about the diagram:

Answer:

$x =$ $°$
$y =$ $°$
$y$ is .

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Angles on a straight line add up to $180°$ .

STEP: <no title>
[−3 points ⇒ 0 / 3 points left]

Line $YZ$ is a straight line, and angles on a straight line up to $180°$ . This means that all of the angles above $YZ$ add up to $180°$ :

\begin{aligned} 35 ° + 82 ° + x & = 180 ° (∠ s on a str line) \\ ∴ x & = 63 ° \end{aligned}

$\begin{aligned} 35° + 82° + x & = 180{°} \quad (\angle \text{s on a str line}) \\ \therefore x & = 63° \end{aligned}$

In the same way, all of the angles below $YZ$ add up to $180°$ :

\begin{aligned} 46 ° + 36 ° + 23 ° + y & = 180 ° (∠ s on a str line) \\ ∴ y & = 75 ° \end{aligned}

$\begin{aligned} 46° + 36° + 23° + y & = 180{°} \quad (\angle \text{s on a str line}) \\ \therefore y & = 75° \end{aligned}$

Here is the completed diagram, with all of the angle values labelled:

The angles in a full turn (revolution) must add up to $360°$ . This is a good way to check our answers:

\begin{aligned} 35 ° + 82 ° + 63 ° + 46 ° + 36 ° + 23 ° + 75 ° & = 360 ° \end{aligned}

$\begin{aligned} 35° + 82° + 63° + 46° + 36° + 23° + 75° & = 360{°} \end{aligned}$

$y = 75°$

$\therefore y$ is an acute angle (less than 90 $°$ ).

Submit your answer as: andand

ID is: 1866 Seed is: 7078

Angles in a full turn

In the diagram below, line $BC$ is a straight line with angles labelled above and below it.

Answer the following questions about the diagram:

Answer:

$x =$ $°$
$y =$ $°$
$y$ is .

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Angles on a straight line add up to $180°$ .

STEP: <no title>
[−3 points ⇒ 0 / 3 points left]

Line $BC$ is a straight line, and angles on a straight line up to $180°$ . This means that all of the angles above $BC$ add up to $180°$ :

\begin{aligned} 40 ° + x + 13 ° & = 180 ° (∠ s on a str line) \\ ∴ x & = 127 ° \end{aligned}

$\begin{aligned} 40° + x + 13° & = 180{°} \quad (\angle \text{s on a str line}) \\ \therefore x & = 127° \end{aligned}$

In the same way, all of the angles below $BC$ add up to $180°$ :

\begin{aligned} 95 ° + 44 ° + 18 ° + y & = 180 ° (∠ s on a str line) \\ ∴ y & = 23 ° \end{aligned}

$\begin{aligned} 95° + 44° + 18° + y & = 180{°} \quad (\angle \text{s on a str line}) \\ \therefore y & = 23° \end{aligned}$

Here is the completed diagram, with all of the angle values labelled:

The angles in a full turn (revolution) must add up to $360°$ . This is a good way to check our answers:

\begin{aligned} 40 ° + 127 ° + 13 ° + 95 ° + 44 ° + 18 ° + 23 ° & = 360 ° \end{aligned}

$\begin{aligned} 40° + 127° + 13° + 95° + 44° + 18° + 23° & = 360{°} \end{aligned}$

$y = 23°$

$\therefore y$ is an acute angle (less than 90 $°$ ).

Submit your answer as: andand

ID is: 4181 Seed is: 2204

Relationships between quadrilateral definitions

It is useful to think about quadrilaterals as a connected family. One type of quadrilateral can have all the same properties as another type of quadrilateral, plus some extra properties.

Select from the options to complete the statement about a square.

Answer: A square is a rhombus with:

shorter sides.
corner angles $\text{90}\text{°}$ .
even sides.
two pairs of opposite sides parallel.

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

What is the definition of a rhombus? What additional properties does a square have?

STEP: Compare the definitions for a square and a rhombus
[−1 point ⇒ 0 / 1 points left]

A rhombus is a quadrilateral with adjacent sides equal and two pairs of opposite sides parallel.

A square is a quadrilateral with two pairs of opposite sides parallel, adjacent sides equal, and all corners

90 °

$\text{90}\text{°}$ .

The following diagram shows a rhombus next to a square, with their defining properties filled in.

We can see that the square has the same properties as the rhombus, plus corner angles $\text{90}\text{°}$ .

Submit your answer as:

corner angles

90 °

$\text{90}\text{°}$ .

ID is: 4181 Seed is: 4747

Relationships between quadrilateral definitions

It is useful to think about quadrilaterals as a connected family. One type of quadrilateral can have all the same properties as another type of quadrilateral, plus some extra properties.

Select from the options to complete the statement about a square.

Answer: A square is a rhombus with:

shorter sides.
even sides.
corner angles $\text{90}\text{°}$ .
two pairs of opposite sides parallel.

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

What is the definition of a rhombus? What additional properties does a square have?

STEP: Compare the definitions for a square and a rhombus
[−1 point ⇒ 0 / 1 points left]

A rhombus is a quadrilateral with two pairs of opposite sides parallel and adjacent sides equal.

A square is a quadrilateral with two pairs of opposite sides parallel, adjacent sides equal, and all corners

90 °

$\text{90}\text{°}$ .

The following diagram shows a rhombus next to a square, with their defining properties filled in.

We can see that the square has the same properties as the rhombus, plus corner angles $\text{90}\text{°}$ .

Submit your answer as:

corner angles

90 °

$\text{90}\text{°}$ .

ID is: 4181 Seed is: 8441

Relationships between quadrilateral definitions

It is useful to think about quadrilaterals as a connected family. One type of quadrilateral can have all the same properties as another type of quadrilateral, plus some extra properties.

Select from the options to complete the statement about a rectangle.

Answer: A rectangle is a parallelogram with:

corner angles $\text{90}\text{°}$ .
two pairs of opposite sides parallel.
two pairs of opposite sides equal.
two pairs of adjacent sides equal.

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

What is the definition of a parallelogram? What additional properties does a rectangle have?

STEP: Compare the definitions for a rectangle and a parallelogram
[−1 point ⇒ 0 / 1 points left]

A parallelogram is a quadrilateral with two pairs of opposite sides parallel.

A rectangle is a quadrilateral with two pairs of opposite sides parallel and all corners

90 °

$\text{90}\text{°}$ .

The following diagram shows a parallelogram next to a rectangle, with their defining properties filled in.

NOTE: It is true that the opposite sides of a rectangle are the same length. But, this is also true for all parallelograms. So, we do not say that a rectangle is a parallelogram with opposite sides equal.

We can see that the rectangle has the same properties as the parallelogram, plus corner angles $\text{90}\text{°}$ .

Submit your answer as:

corner angles

90 °

$\text{90}\text{°}$ .

ID is: 4300 Seed is: 8349

Using diagonals to prove properties

In the diagram below, $WVXY$ is a quadrilateral with $WV = VX$ and $\angle WVZ = \angle XVZ = \text{39}\text{°}$ .

How should we prove that $W\hat YZ = X\hat YZ$ ?
Answer:
Prove that $\Delta WXY \equiv \Delta XWV$ .
Deduce that $W\hat YZ = X\hat YZ$ because diagonals bisect vertices.
Prove that $\Delta WVZ \equiv \Delta XVZ$ .
Prove that $\Delta WVY \equiv \Delta XVY$ .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You must only use the information you have been given. Can you use this information to prove that any triangles are congruent? Then, their matching angles and sides will be equal.

STEP: Choose which triangles to prove congruent
[−1 point ⇒ 0 / 1 points left]

$W\hat YZ$ is a angle in $\Delta WVY$ and $X\hat YZ$ is the matching angle in $\Delta XVY$ . So, if we prove that $\Delta WVY \equiv \Delta XVY$ , then $W\hat YZ$ and $X\hat YZ$ must be equal because congruent triangles are equal in every way.

We also have to check that we have enough information to prove that $\Delta WVY \equiv \Delta XVY$ . See if you can spot two pairs of sides and one pair of angles that are equal between the two triangles!

Submit your answer as:
Prove that $\Delta WVY \equiv \Delta XVY$ .
The diagram of quadrilateral $WVXY$ is repeated for convenience:

Now, prove that $W\hat YZ = X\hat YZ$ by selecting the correct options.
Answer:
In :
1. is common.
3. $\angle WVZ = \angle XVZ$ (given)
$\therefore$

$\therefore W\hat YZ = X\hat YZ$
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Prove congruency by matching up pairs of sides and angles. Use the given information and geometry facts.

STEP: Complete the proof
[−6 points ⇒ 0 / 6 points left]
TIP:
- Look for any sides or angles that the two triangles have in common.
- Then, write down any pairs of equal sides or angles that you have been given.
- Finally, decide what else you need in order to prove that the two triangles are congruent, according to one of the four cases of congruency.
The correct proof is:

In $\Delta WVY$ and $\Delta XVY$ :
1. $VY$ is common.
2. $WV = VX$ $\left(\text{given}\right)$
3. $\angle WVZ = \angle XVZ$ $\left(\text{given}\right)$
$\therefore \Delta WVY \equiv \Delta XVY$ $\left(\text{SAS}\right)$

$\therefore W\hat YZ = X\hat YZ$ $(\Delta WVY \equiv \Delta XVY)$

NOTE: You can extend this proof to prove that the diagonals of any rhombus bisect all of its vertices.

Submit your answer as: andandandandand

ID is: 4300 Seed is: 417

Using diagonals to prove properties

In the diagram below, $WVXY$ is a quadrilateral with $WV = VX$ and $\angle WVZ = \angle XVZ = \text{24}\text{°}$ .

How should we prove that $W\hat YZ = X\hat YZ$ ?
Answer:
Prove that $\Delta WVZ \equiv \Delta XVZ$ .
Prove that $\Delta WXY \equiv \Delta XWV$ .
Deduce that $W\hat YZ = X\hat YZ$ because diagonals bisect vertices.
Prove that $\Delta WVY \equiv \Delta XVY$ .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You must only use the information you have been given. Can you use this information to prove that any triangles are congruent? Then, their matching angles and sides will be equal.

STEP: Choose which triangles to prove congruent
[−1 point ⇒ 0 / 1 points left]

$W\hat YZ$ is a angle in $\Delta WVY$ and $X\hat YZ$ is the matching angle in $\Delta XVY$ . So, if we prove that $\Delta WVY \equiv \Delta XVY$ , then $W\hat YZ$ and $X\hat YZ$ must be equal because congruent triangles are equal in every way.

We also have to check that we have enough information to prove that $\Delta WVY \equiv \Delta XVY$ . See if you can spot two pairs of sides and one pair of angles that are equal between the two triangles!

Submit your answer as:
Prove that $\Delta WVY \equiv \Delta XVY$ .
The diagram of quadrilateral $WVXY$ is repeated for convenience:

Now, prove that $W\hat YZ = X\hat YZ$ by selecting the correct options.
Answer:
In :
1. is common.
3. $\angle WVZ = \angle XVZ$ (given)
$\therefore$

$\therefore W\hat YZ = X\hat YZ$
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Prove congruency by matching up pairs of sides and angles. Use the given information and geometry facts.

STEP: Complete the proof
[−6 points ⇒ 0 / 6 points left]
TIP:
- Look for any sides or angles that the two triangles have in common.
- Then, write down any pairs of equal sides or angles that you have been given.
- Finally, decide what else you need in order to prove that the two triangles are congruent, according to one of the four cases of congruency.
The correct proof is:

In $\Delta WVY$ and $\Delta XVY$ :
1. $VY$ is common.
2. $WV = VX$ $\left(\text{given}\right)$
3. $\angle WVZ = \angle XVZ$ $\left(\text{given}\right)$
$\therefore \Delta WVY \equiv \Delta XVY$ $\left(\text{SAS}\right)$

$\therefore W\hat YZ = X\hat YZ$ $(\Delta WVY \equiv \Delta XVY)$

NOTE: You can extend this proof to prove that the diagonals of any rhombus bisect all of its vertices.

Submit your answer as: andandandandand

ID is: 4300 Seed is: 8103

Using diagonals to prove properties

In the diagram below, $WVXY$ is a quadrilateral with $WV = VX$ and $\angle WVZ = \angle XVZ = \text{29}\text{°}$ .

How should we prove that $W\hat YZ = X\hat YZ$ ?
Answer:
Prove that $\Delta WXY \equiv \Delta XWV$ .
Deduce that $W\hat YZ = X\hat YZ$ because diagonals bisect vertices.
Prove that $\Delta WVY \equiv \Delta XVY$ .
Prove that $\Delta WVZ \equiv \Delta XVZ$ .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You must only use the information you have been given. Can you use this information to prove that any triangles are congruent? Then, their matching angles and sides will be equal.

STEP: Choose which triangles to prove congruent
[−1 point ⇒ 0 / 1 points left]

$W\hat YZ$ is a angle in $\Delta WVY$ and $X\hat YZ$ is the matching angle in $\Delta XVY$ . So, if we prove that $\Delta WVY \equiv \Delta XVY$ , then $W\hat YZ$ and $X\hat YZ$ must be equal because congruent triangles are equal in every way.

We also have to check that we have enough information to prove that $\Delta WVY \equiv \Delta XVY$ . See if you can spot two pairs of sides and one pair of angles that are equal between the two triangles!

Submit your answer as:
Prove that $\Delta WVY \equiv \Delta XVY$ .
The diagram of quadrilateral $WVXY$ is repeated for convenience:

Now, prove that $W\hat YZ = X\hat YZ$ by selecting the correct options.
Answer:
In :
1. is common.
3. $\angle WVZ = \angle XVZ$ (given)
$\therefore$

$\therefore W\hat YZ = X\hat YZ$
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Prove congruency by matching up pairs of sides and angles. Use the given information and geometry facts.

STEP: Complete the proof
[−6 points ⇒ 0 / 6 points left]
TIP:
- Look for any sides or angles that the two triangles have in common.
- Then, write down any pairs of equal sides or angles that you have been given.
- Finally, decide what else you need in order to prove that the two triangles are congruent, according to one of the four cases of congruency.
The correct proof is:

In $\Delta WVY$ and $\Delta XVY$ :
1. $VY$ is common.
2. $WV = VX$ $\left(\text{given}\right)$
3. $\angle WVZ = \angle XVZ$ $\left(\text{given}\right)$
$\therefore \Delta WVY \equiv \Delta XVY$ $\left(\text{SAS}\right)$

$\therefore W\hat YZ = X\hat YZ$ $(\Delta WVY \equiv \Delta XVY)$

NOTE: You can extend this proof to prove that the diagonals of any rhombus bisect all of its vertices.

Submit your answer as: andandandandand

ID is: 1660 Seed is: 5416

Proof: is the quadrilateral a parallelogram?

The figure below shows quadrilateral $QRST$ with $\angle Q = \angle S = 59°$ and $\angle R = \angle T = 121°$ .

The steps and reasons below prove that $QRST$ is a parallelogram. However, there are two steps and two reasons missing. Choose the correct steps and reasons to complete the proof.

Answer:

STEPS	REASONS
	Given $\left(59° + 121° = 180° \right)$
$QR \parallel TS$
$\angle T + \angle S = 180°$
	Co-interior angles of parallel lines have a sum of $180°$
$QRST$ is a parallelogram	Definition of a parallelogram

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

You need to start at the top, and go down the proof line by line. Work out the answers as you go down the proof. And stay patient: this question requires a lot of careful thinking!

STEP: Consider the options and choose the correct steps and reasons
[−4 points ⇒ 0 / 4 points left]

The definition of a parallelogram is: "a quadrilateral with both pairs of opposite sides parallel." To prove that the quadrilateral is a parallelogram, we need to show that the opposite sides are parallel to each other. In other words, the proof should follow whatever steps are needed to show that side $QR$ is parallel to side $TS$ and that side $RS$ is parallel to side $QT$ .

This proof uses the angle values to show that adjacent angles must be co-interior angles within parallel lines. The table below explains the various steps of the proof, using colour to tie the explanation to the steps and reasons in the proof.

$\quad$ Explanations

$\quad$

The yellow rows in the table indicate the given information. This is information we can gather directly from the diagram. The size of all the angles are given.

The green rows represent the application of co-interior angles. Co-interior angles inside parallel lines are supplementary (they have a sum of $180°$ ). Here we use this fact in reverse: if the angles are supplementary the lines are parallel.

Once we have shown that the opposite sides are parallel to each other, we can conclude that quadrilateral

Q R S T

$QRST$ is a parallelogram. Note that the final step of a proof will always be the fact that you are trying to prove.

Here is the completed proof with the correct steps and reasons.

$\quad$ Steps $\quad$	Reasons
$\angle Q + \angle T = 180°$	Given $\left(59° + 121° = 180° \right)$
$QR \parallel TS$	Co-interior angles of parallel lines have a sum of $180°$
$\angle T + \angle S = 180°$	Given $\left( 121° + 59° = 180° \right)$
$RS \parallel QT$	Co-interior angles of parallel lines have a sum of $180°$
$QRST$ is a parallelogram	Definition of a parallelogram

Submit your answer as: andandand

ID is: 1660 Seed is: 2940

Proof: is the quadrilateral a parallelogram?

The figure below shows quadrilateral $XWVU$ with $\angle X = \angle V = 74°$ and $\angle W = \angle U = 106°$ .

The steps and reasons below prove that $XWVU$ is a parallelogram. However, there are two steps and two reasons missing. Choose the correct steps and reasons to complete the proof.

Answer:

STEPS	REASONS
	Given $\left(74° + 106° = 180° \right)$
	Co-interior angles of parallel lines have a sum of $180°$
$\angle U + \angle V = 180°$
$WV \parallel XU$
$XWVU$ is a parallelogram	Definition of a parallelogram

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

You need to start at the top, and go down the proof line by line. Work out the answers as you go down the proof. And stay patient: this question requires a lot of careful thinking!

STEP: Consider the options and choose the correct steps and reasons
[−4 points ⇒ 0 / 4 points left]

The definition of a parallelogram is: "a quadrilateral with both pairs of opposite sides parallel." To prove that the quadrilateral is a parallelogram, we need to show that the opposite sides are parallel to each other. In other words, the proof should follow whatever steps are needed to show that side $XW$ is parallel to side $UV$ and that side $WV$ is parallel to side $XU$ .

$\quad$ Explanations

$\quad$

The yellow rows in the table indicate the given information. This is information we can gather directly from the diagram. The size of all the angles are given.

Once we have shown that the opposite sides are parallel to each other, we can conclude that quadrilateral

X W V U

$XWVU$ is a parallelogram. Note that the final step of a proof will always be the fact that you are trying to prove.

Here is the completed proof with the correct steps and reasons.

$\quad$ Steps $\quad$	Reasons
$\angle X + \angle U = 180°$	Given $\left(74° + 106° = 180° \right)$
$XW \parallel UV$	Co-interior angles of parallel lines have a sum of $180°$
$\angle U + \angle V = 180°$	Given $\left( 106° + 74° = 180° \right)$
$WV \parallel XU$	Co-interior angles of parallel lines have a sum of $180°$
$XWVU$ is a parallelogram	Definition of a parallelogram

Submit your answer as: andandand

ID is: 1660 Seed is: 8838

Proof: is the quadrilateral a parallelogram?

The figure below shows quadrilateral $ABCD$ with $\angle A = \angle C = 121°$ and $\angle B = \angle D = 59°$ .

The steps and reasons below prove that $ABCD$ is a parallelogram. However, there are two steps and two reasons missing. Choose the correct steps and reasons to complete the proof.

Answer:

STEPS	REASONS
	Given $\left(121° + 59° = 180° \right)$
$AB \parallel DC$	Co-interior angles of parallel lines have a sum of $180°$
$\angle D + \angle C = 180°$
	Co-interior angles of parallel lines have a sum of $180°$
$ABCD$ is a parallelogram

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

You need to start at the top, and go down the proof line by line. Work out the answers as you go down the proof. And stay patient: this question requires a lot of careful thinking!

STEP: Consider the options and choose the correct steps and reasons
[−4 points ⇒ 0 / 4 points left]

The definition of a parallelogram is: "a quadrilateral with both pairs of opposite sides parallel." To prove that the quadrilateral is a parallelogram, we need to show that the opposite sides are parallel to each other. In other words, the proof should follow whatever steps are needed to show that side $AB$ is parallel to side $DC$ and that side $BC$ is parallel to side $AD$ .

$\quad$ Explanations

$\quad$

The yellow rows in the table indicate the given information. This is information we can gather directly from the diagram. The size of all the angles are given.

Once we have shown that the opposite sides are parallel to each other, we can conclude that quadrilateral

A B C D

$ABCD$ is a parallelogram. Note that the final step of a proof will always be the fact that you are trying to prove.

Here is the completed proof with the correct steps and reasons.

$\quad$ Steps $\quad$	Reasons
$\angle A + \angle D = 180°$	Given $\left(121° + 59° = 180° \right)$
$AB \parallel DC$	Co-interior angles of parallel lines have a sum of $180°$
$\angle D + \angle C = 180°$	Given $\left( 59° + 121° = 180° \right)$
$BC \parallel AD$	Co-interior angles of parallel lines have a sum of $180°$
$ABCD$ is a parallelogram	Definition of a parallelogram

Submit your answer as: andandand

ID is: 4215 Seed is: 4463

Relationships between quadrilaterals: true or false

Consider the following statement and decide whether it is true or false.

All kites are squares.

Answer: The statement is .

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Think about the definitions of the quadrilaterals and how they relate to each other. Does a kite always have at least the same properties as a square? Or, can you think of any kites that are not types of squares?

STEP: Think about the definitions of the quadrilaterals
[−2 points ⇒ 0 / 2 points left]

A kite is a quadrilateral with two pairs of equal adjacent sides.

A square is a quadrilateral with two pairs of opposite sides that are parallel, equal adjacent sides, and corners that are

90 °

$\text{90}\text{°}$ .

Consider the following kite:

This shape meets the definition of a kite. But, it does not meet the definition of a square because the opposite sides are not parallel (amongst other things). So, we have come up with an example that disagrees with the statement. Therefore, the statement is false.

NOTE: An example that disagrees with the statement is called a counter-example. We only need one counter-example to prove that a statement is false. This is because if it is false for one example, then we cannot say that it is true for all - in other words, it is false.

Submit your answer as:

ID is: 4215 Seed is: 37

Relationships between quadrilaterals: true or false

Consider the following statement and decide whether it is true or false.

All squares are rhombuses.

Answer: The statement is .

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Think about the definitions of the quadrilaterals and how they relate to each other. Does a square always have at least the same properties as a rhombus? Or, can you think of any squares that are not types of rhombuses?

STEP: Think about the definitions of the quadrilaterals
[−2 points ⇒ 0 / 2 points left]

A square is a quadrilateral with two pairs of opposite sides that are parallel, equal adjacent sides, and corners that are

90 °

$\text{90}\text{°}$ .

A rhombus is a quadrilateral with two pairs of opposite sides that are parallel and equal adjacent sides.

We can see that a square has all of the same properties as a rhombus (and some more). So, squares are special types of rhombuses. Therefore, the statement is true.

NOTE: It does not matter that a square has extra properties that a rhombus does not have. What matters is that it has at least the properties of a rhombus.

Submit your answer as:

ID is: 4215 Seed is: 141

Relationships between quadrilaterals: true or false

Consider the following statement and decide whether it is true or false.

All squares are kites.

Answer: The statement is .

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Think about the definitions of the quadrilaterals and how they relate to each other. Does a square always have at least the same properties as a kite? Or, can you think of any squares that are not types of kites?

STEP: Think about the definitions of the quadrilaterals
[−2 points ⇒ 0 / 2 points left]

A square is a quadrilateral with two pairs of opposite sides that are parallel, equal adjacent sides, and corners that are

90 °

$\text{90}\text{°}$ .

A kite is a quadrilateral with two pairs of equal adjacent sides.

We can see that a square has all of the same properties as a kite (and some more). So, squares are special types of kites. Therefore, the statement is true.

NOTE: It does not matter that a square has extra properties that a kite does not have. What matters is that it has at least the properties of a kite.

Submit your answer as:

ID is: 4299 Seed is: 7598

Proving diagonals using congruency

In the diagram below, $WVXY$ is a quadrilateral with $WV = XY$ and $WV \parallel XY$ .

Prove that $ZW = ZX$ .

How should we prove that $ZW = ZX$ ?
Answer:
Prove that $\Delta YZX \equiv \Delta VZW$ .
Deduce that $ZW = ZX$ because $\hat Z$ is common.
Deduce that $ZW = ZX$ because diagonals of a parallelogram bisect.
Prove that $\Delta WVY \equiv \Delta XYV$ .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You must only use the information you have been given. Can you use this information to prove that any triangles are congruent? Then, their matching angles and sides will be equal.

STEP: Choose which triangles to prove congruent
[−1 point ⇒ 0 / 1 points left]

The correct strategy is: Prove that $\Delta YZX \equiv \Delta VZW$ .

$ZW$ is a side in $\Delta VZW$ . $ZX$ is the matching side in $\Delta YZX$ . So, if we prove that $\Delta YZX \equiv \Delta VZX$ , then $ZW$ and $ZX$ must be equal because congruent triangles are equal in every way.

We also have to check that we have enough information to prove that $\Delta YZX \equiv \Delta VZX$ . See if you can spot one pair of sides and two pairs of angles that are equal between the two triangles!

Submit your answer as:
Prove that $\Delta YZX \equiv \Delta VZW$ .
The diagram of quadrilateral $WVXY$ is repeated for convenience:

Now, prove that $ZW = ZX$ by selecting the correct options.
Answer:
In :
1. $WV = XY$ (given)
2. $\angle ZVW =$
3. $\angle ZWV =$
$\therefore$ (SAA)
$\therefore ZW = ZX$
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Prove congruency by matching up pairs of sides and angles. Use the given information and geometry facts.

STEP: Complete the proof
[−6 points ⇒ 0 / 6 points left]
TIP:
- Look for any sides or angles that the two triangles have in common.
- Then, write down any pairs of equal sides or angles that you have been given.
- Finally, decide what else you need in order to prove that the two triangles are congruent, according to one of the four cases of congruency.
The correct proof is:

In $\Delta YZX$ and $\Delta VZW$ :
1. $WV = XY$ $\left(\text{given}\right)$
2. $\angle ZVW = \angle ZYX$ $\left(\text{alt } \angle \text{s; } WV \parallel XY\right)$
3. $\angle ZWV = \angle ZXY$ $\left(\text{alt } \angle \text{s; } WV \parallel XY\right)$
$\therefore$ $\Delta YZX \equiv \Delta VZW$ $\left(\text{SAA}\right)$

$\therefore$ $ZW = ZX$ $\left(\Delta YZX \equiv \Delta VZW\right)$

NOTE: You can extend this proof to prove that diagonals bisect each other in all parallelograms. This is one of the properties that you should know already. Now you know why it is true!

Submit your answer as: andandandandandand

ID is: 4299 Seed is: 9718

Proving diagonals using congruency

In the diagram below, $ABCD$ is a quadrilateral with $AB = CD$ and $AB \parallel CD$ .

Prove that $EA = EC$ .

How should we prove that $EA = EC$ ?
Answer:
Deduce that $EA = EC$ because diagonals of a parallelogram bisect.
Deduce that $EA = EC$ because $\hat E$ is common.
Prove that $\Delta DEC \equiv \Delta BEA$ .
Prove that $\Delta ABD \equiv\Delta BAC$ .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You must only use the information you have been given. Can you use this information to prove that any triangles are congruent? Then, their matching angles and sides will be equal.

STEP: Choose which triangles to prove congruent
[−1 point ⇒ 0 / 1 points left]

The correct strategy is: Prove that $\Delta DEC \equiv \Delta BEA$ .

$EA$ is a side in $\Delta BEA$ . $EC$ is the matching side in $\Delta DEC$ . So, if we prove that $\Delta DEC \equiv \Delta BEC$ , then $EA$ and $EC$ must be equal because congruent triangles are equal in every way.

We also have to check that we have enough information to prove that $\Delta DEC \equiv \Delta BEC$ . See if you can spot one pair of sides and two pairs of angles that are equal between the two triangles!

Submit your answer as:
Prove that $\Delta DEC \equiv \Delta BEA$ .
The diagram of quadrilateral $ABCD$ is repeated for convenience:

Now, prove that $EA = EC$ by selecting the correct options.
Answer:
In :
1. $AB = CD$ (given)
2. $\angle EBA =$
3. $\angle EAB =$
$\therefore$ (SAA)
$\therefore EA = EC$
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Prove congruency by matching up pairs of sides and angles. Use the given information and geometry facts.

STEP: Complete the proof
[−6 points ⇒ 0 / 6 points left]
TIP:
- Look for any sides or angles that the two triangles have in common.
- Then, write down any pairs of equal sides or angles that you have been given.
- Finally, decide what else you need in order to prove that the two triangles are congruent, according to one of the four cases of congruency.
The correct proof is:

In $\Delta DEC$ and $\Delta BEA$ :
1. $AB = CD$ $\left(\text{given}\right)$
2. $\angle EBA = \angle EDC$ $\left(\text{alt } \angle \text{s; } AB \parallel CD\right)$
3. $\angle EAB = \angle ECD$ $\left(\text{alt } \angle \text{s; } AB \parallel CD\right)$
$\therefore$ $\Delta DEC \equiv \Delta BEA$ $\left(\text{SAA}\right)$

$\therefore$ $EA = EC$ $\left(\Delta DEC \equiv \Delta BEA\right)$

NOTE: You can extend this proof to prove that diagonals bisect each other in all parallelograms. This is one of the properties that you should know already. Now you know why it is true!

Submit your answer as: andandandandandand

ID is: 4299 Seed is: 8785

Proving diagonals using congruency

In the diagram below, $PQRS$ is a quadrilateral with $PS = QR$ and $P\hat QR = Q\hat PS =\:$ $\text{90}\text{°}$ .

Prove that $PR = QS$ .

How should we prove that $PR = QS$ ?
Answer:
Deduce that $PR = QS$ because $\hat T$ is common.
Prove that $\Delta PQS \equiv \Delta QPR$ .
Prove that $\Delta PQS \equiv \Delta RSQ$ .
Prove that $\Delta STP \equiv \Delta QTR$ .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You must only use the information you have been given. Can you use this information to prove that any triangles are congruent? Then, their matching angles and sides will be equal.

STEP: Choose which triangles to prove congruent
[−1 point ⇒ 0 / 1 points left]

The correct strategy is: Prove that $\Delta PQS \equiv \Delta QPR$ .

$PR$ is a side in $\Delta QPR$ and $QS$ is the matching side in $\Delta PQS$ . So, if we prove that $\Delta PQS \equiv \Delta QPR$ , then $PR$ and $QS$ must be equal because congruent triangles are equal in every way.

We also have to check that we have enough information to prove that $\Delta PQS \equiv \Delta QPR$ . See if you can spot two pairs of sides and one pair of angles that are equal between the two triangles!

Submit your answer as:
Prove that $\Delta PQS \equiv \Delta QPR$ .
The diagram of quadrilateral $PQRS$ is repeated for convenience:

Now, prove that $PR = QS$ by selecting the correct options.
Answer:
In :
1. is common.
2. $PS = QR$
3. $\angle PQR = \angle SPQ = \text{90}\text{°}$ (given)
$\therefore$
$\therefore PR = QS$
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Prove congruency by matching up pairs of sides and angles. Use the given information and geometry facts.

STEP: Complete the proof
[−6 points ⇒ 0 / 6 points left]
TIP:
- Look for any sides or angles that the two triangles have in common.
- Then, write down any pairs of equal sides or angles that you have been given.
- Finally, decide what else you need in order to prove that the two triangles are congruent, according to one of the four cases of congruency.
The correct proof is:

In $\Delta PQS$ and $\Delta QPR$ :
1. $PQ$ is common.
2. $PS = QR$ $\left(\text{given}\right)$
3. $\angle PQR = \angle SPQ = \text{90}\text{°}$ $\left(\text{given}\right)$
$\therefore$ $\Delta PQS \equiv \Delta QPR$ $\left(\text{SAS}\right)$

$\therefore$ $PR = QS$ $\left(\Delta PQS \equiv \Delta QPR\right)$

NOTE: You can extend this proof to prove that diagonals are equal in all rectangles. This is one of the properties that you should know already. Now you know why it is true!

Submit your answer as: andandandandand

ID is: 1862 Seed is: 1427

Calculate angles between parallel lines

In the diagram below, $FG \parallel$ $HJ$ . $KL$ is a transversal.

NOTE: A transversal is a line that cuts across two other lines.

What is the value of $x$ ?

Answer:

x =

$x =$

°

${°}$

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

STEP: <no title>
[−2 points ⇒ 0 / 2 points left]

Corresponding angles are in the same position compared to the parallel lines and the transversal. If we highlight the angles we will see an "F" shape. This helps us to identify corresponding angles.

Since we know that corresponding angles are equal, we can write:

\begin{aligned} x & = 104 ° (corresp ∠ s, F G ∥ H J) \end{aligned}

$\begin{aligned} x &= 104° \quad (\text{corresp } \angle\text{s}, FG \parallel HJ ) \end{aligned}$

Therefore $x = 104°$ (corresp $\angle$ s, $FG \parallel HJ$ ).

NOTE: Although you didn't need to give a reason here, reasons are important in geometry. You should learn these reasons and always write them in your work.

Submit your answer as:

ID is: 1862 Seed is: 6355

Calculate angles between parallel lines

In the diagram below, $CD \parallel$ $EF$ . $GH$ is a transversal.

NOTE: A transversal is a line that cuts across two other lines.

What is the value of $x$ ?

Answer:

x =

$x =$

°

${°}$

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

STEP: <no title>
[−2 points ⇒ 0 / 2 points left]

Alternate angles are both inside the parallel lines, but on opposite sides of the transversal. If we highlight the angles we will see a "Z" shape. This helps us to identify alternate angles.

Since we know that alternate angles are equal, we can write:

\begin{aligned} x & = 50 ° (alt ∠ s, C D ∥ E F) \end{aligned}

$\begin{aligned} x &= 50° \quad (\text{alt } \angle\text{s}, CD \parallel EF ) \end{aligned}$

Therefore $x = 50°$ (alt $\angle$ s, $CD \parallel EF$ ).

NOTE: Although you didn't need to give a reason here, reasons are important in geometry. You should learn these reasons and always write them in your work.

Submit your answer as:

ID is: 1862 Seed is: 5441

Calculate angles between parallel lines

In the diagram below, $DE \parallel$ $FG$ . $HJ$ is a transversal.

NOTE: A transversal is a line that cuts across two other lines.

What is the value of $x$ ?

Answer:

x =

$x =$

°

${°}$

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

STEP: <no title>
[−2 points ⇒ 0 / 2 points left]

Alternate angles are both inside the parallel lines, but on opposite sides of the transversal. If we highlight the angles we will see a "Z" shape. This helps us to identify alternate angles.

Since we know that alternate angles are equal, we can write:

\begin{aligned} x & = 96 ° (alt ∠ s, D E ∥ F G) \end{aligned}

$\begin{aligned} x &= 96° \quad (\text{alt } \angle\text{s}, DE \parallel FG ) \end{aligned}$

Therefore $x = 96°$ (alt $\angle$ s, $DE \parallel FG$ ).

NOTE: Although you didn't need to give a reason here, reasons are important in geometry. You should learn these reasons and always write them in your work.

Submit your answer as:

ID is: 4252 Seed is: 7898

Given diagonal properties, identify the quadrilateral

Consider the following information about the diagonals of a quadrilateral:

The diagonals are perpendicular to each other.

Answer:

The statement is definitely true for all:

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Think about the diagonal properties of the different quadrilaterals. Which one has the exact properties described in the question?

STEP: Recall diagonal properties
[−2 points ⇒ 0 / 2 points left]

Perpendicular means that the diagonals meet at $\text{90}\text{°}$ . This is true for all kites.

NOTE: This is true for all kites, so it must also be true for rhombuses and squares (which are types of kites).

Submit your answer as:

kites

ID is: 4252 Seed is: 1311

Given diagonal properties, identify the quadrilateral

Consider the following information about the diagonals of a quadrilateral:

The diagonals are perpendicular bisectors of each other.

Answer:

The statement is definitely true for all:

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Think about the diagonal properties of the different quadrilaterals. Which one has the exact properties described in the question?

STEP: Recall diagonal properties
[−2 points ⇒ 0 / 2 points left]

Perpendicular means that the diagonals meet at $\text{90}\text{°}$ . This is true for all kites.

The diagonals of all parallelograms bisect each other.

NOTE: Bisect means to cut into two equal parts.

So, quadrilaterals that have these properties must be types of kites, and also types of parallelograms. So, the quadrilaterals for which this is definitely true are rhombuses.

NOTE: This is true for all rhombuses, so it must also be true for squares (which are types of rhombuses).

Submit your answer as:

rhombuses

ID is: 4252 Seed is: 6937

Given diagonal properties, identify the quadrilateral

Consider the following information about the diagonals of a quadrilateral:

The diagonals are equal in length.

Answer:

The statement is definitely true for all:

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Think about the diagonal properties of the different quadrilaterals. Which one has the exact properties described in the question?

STEP: Recall diagonal properties
[−2 points ⇒ 0 / 2 points left]

The diagonals of all rectangles are equal in length. So, the quadrilateral must be a rectangle.

NOTE: This is true for all rectangles, so it must also be true for squares (which are types of rectangles).

Submit your answer as:

rectangles

ID is: 1867 Seed is: 1142

Recognise pairs of angles between parallel lines

In the diagram below, ${AB} \parallel {CD}$ . $\ {EF}$ is a transversal.

NOTE: A transversal is a line that cuts across two other lines.

What type of angle pair is shown by the coloured angles?

Answer: The angles are .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Can you identify the relationship between the coloured angles?

STEP: Identify the type of angle pair
[−1 point ⇒ 0 / 1 points left]

This intersection creates corresponding angles.

Submit your answer as:

ID is: 1867 Seed is: 5144

Recognise pairs of angles between parallel lines

In the diagram below, ${AB} \parallel {CD}$ . $\ {EF}$ is a transversal.

NOTE: A transversal is a line that cuts across two other lines.

What type of angle pair is shown by the coloured angles?

Answer: The angles are .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Can you identify the relationship between the coloured angles?

STEP: Identify the type of angle pair
[−1 point ⇒ 0 / 1 points left]

This intersection creates co-interior angles.

Co-interior angles are both inside the parallel lines, on the same side of the transversal. If we highlight the angles we will see a "U" shape. This helps us identify co-interior angles.

Submit your answer as:

ID is: 1867 Seed is: 7432

Recognise pairs of angles between parallel lines

In the diagram below, ${AB} \parallel {CD}$ . $\ {EF}$ is a transversal.

NOTE: A transversal is a line that cuts across two other lines.

What type of angle pair is shown by the coloured angles?

Answer: The angles are .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Can you identify the relationship between the coloured angles?

STEP: Identify the type of angle pair
[−1 point ⇒ 0 / 1 points left]

This intersection creates co-interior angles.

Co-interior angles are both inside the parallel lines, on the same side of the transversal. If we highlight the angles we will see a "U" shape. This helps us identify co-interior angles.

Submit your answer as:

ID is: 4276 Seed is: 6678

Congruency in kites

Consider the following diagram. $ABCD$ is a kite.

Prove that $\Delta ABE\equiv \Delta ADE$ by completing the proof below.
Answer:
In $\Delta ABE$ and $\Delta ADE$ :
1. $AB =$
2. $= DE$
3. is common
$\therefore\Delta ABE\equiv \Delta ADE$
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Use the properties of a kite to work out which sides are equal to each other.

STEP: Use the properties of a kite to match up equal sides
[−6 points ⇒ 0 / 6 points left]

First, we focus on the two triangles that we need to prove congruent, and ignore the rest of the diagram.

A kite is a quadrilateral that has two pairs of adjacent sides equal. Using this, and the information we were given in the diagram, we will highlight the sides which we know must be equal in our two triangles.

In $\Delta ABE$ and $\Delta ADE$ :
1. $\color{magenta}{AB = AD}$ (adjacent sides of kite)
2. $\color{ForestGreen}{BE = DE}$ (given)
3. $\color{blue}{AE}$ is common
We have proved that three pairs of sides are equal, so we can use the congruency case SSS.

$\therefore \Delta ABE\equiv \Delta ADE$ (SSS)

Submit your answer as: andandandandand
Consider the following diagram. $ABCD$ is a kite.

You have just proved that $\Delta ABE\equiv \Delta ADE$ . Hence, determine the size of $A\hat DE$ .

Answer: $A\hat DE =$ $\text{°}$
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Once you have proved congruency, you know that matching angles will be equal. Which angle matches $A\hat DE$ ?

STEP: Identify matching angles to determine size
[−2 points ⇒ 0 / 2 points left]

We have proved that $\Delta ABE\equiv \Delta ADE$ .

Therefore, the matching angles in these two triangles must be equal.

In particular, $A\hat DE = A\hat BE$ .

They are only equal because the triangles are congruent, so we use the congruency statement as our reason.

$\therefore A\hat DE =$ $\text{40}\text{°}$ $(\Delta ABE\equiv \Delta ADE)$

Submit your answer as: and

ID is: 4276 Seed is: 6046

Congruency in kites

Consider the following diagram. $PQRS$ is a kite.

Prove that $\Delta PQR\equiv \Delta PSR$ by completing the proof below.
Answer:
In $\Delta PQR$ and $\Delta PSR$ :
1. $PQ =$
2. $= SR$
3. is common
$\therefore\Delta PQR\equiv \Delta PSR$
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Use the properties of a kite to work out which sides are equal to each other.

STEP: Use the properties of a kite to match up equal sides
[−6 points ⇒ 0 / 6 points left]

First, we focus on the two triangles that we need to prove congruent, and ignore the rest of the diagram.

A kite is a quadrilateral that has two pairs of adjacent sides equal. Using this, and the information we were given in the diagram, we will highlight the sides which we know must be equal in our two triangles.

In $\Delta PQR$ and $\Delta PSR$ :
1. $\color{magenta}{PQ = PS}$ (adjacent sides of kite)
2. $\color{ForestGreen}{QR = SR}$ (adjacent sides of kite)
3. $\color{blue}{PR}$ is common
We have proved that three pairs of sides are equal, so we can use the congruency case SSS.

$\therefore \Delta PQR\equiv \Delta PSR$ (SSS)

Submit your answer as: andandandandand
Consider the following diagram. $PQRS$ is a kite.

You have just proved that $\Delta PQR\equiv \Delta PSR$ . Hence, determine the size of $P\hat RS$ .

Answer: $P\hat RS =$ $\text{°}$
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Once you have proved congruency, you know that matching angles will be equal. Which angle matches $P\hat RS$ ?

STEP: Identify matching angles to determine size
[−2 points ⇒ 0 / 2 points left]

We have proved that $\Delta PQR\equiv \Delta PSR$ .

Therefore, the matching angles in these two triangles must be equal.

In particular, $P\hat RS = Q\hat RP$ .

They are only equal because the triangles are congruent, so we use the congruency statement as our reason.

$\therefore P\hat RS =$ $\text{23}\text{°}$ $(\Delta PQR\equiv \Delta PSR)$

Submit your answer as: and

ID is: 4276 Seed is: 4410

Congruency in kites

Consider the following diagram. $CBAD$ is a kite.

Prove that $\Delta CBE\equiv \Delta CDE$ by completing the proof below.
Answer:
In $\Delta CBE$ and $\Delta CDE$ :
1. $CB =$
2. $= DE$
3. is common
$\therefore\Delta CBE\equiv \Delta CDE$
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Use the properties of a kite to work out which sides are equal to each other.

STEP: Use the properties of a kite to match up equal sides
[−6 points ⇒ 0 / 6 points left]

First, we focus on the two triangles that we need to prove congruent, and ignore the rest of the diagram.

A kite is a quadrilateral that has two pairs of adjacent sides equal. Using this, and the information we were given in the diagram, we will highlight the sides which we know must be equal in our two triangles.

In $\Delta CBE$ and $\Delta CDE$ :
1. $\color{magenta}{CB = CD}$ (adjacent sides of kite)
2. $\color{ForestGreen}{BE = DE}$ (given)
3. $\color{blue}{CE}$ is common
We have proved that three pairs of sides are equal, so we can use the congruency case SSS.

$\therefore \Delta CBE\equiv \Delta CDE$ (SSS)

Submit your answer as: andandandandand
Consider the following diagram. $CBAD$ is a kite.

You have just proved that $\Delta CBE\equiv \Delta CDE$ . Hence, determine the size of $C\hat DE$ .

Answer: $C\hat DE =$ $\text{°}$
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Once you have proved congruency, you know that matching angles will be equal. Which angle matches $C\hat DE$ ?

STEP: Identify matching angles to determine size
[−2 points ⇒ 0 / 2 points left]

We have proved that $\Delta CBE\equiv \Delta CDE$ .

Therefore, the matching angles in these two triangles must be equal.

In particular, $C\hat DE = C\hat BE$ .

They are only equal because the triangles are congruent, so we use the congruency statement as our reason.

$\therefore C\hat DE =$ $\text{42}\text{°}$ $(\Delta CBE\equiv \Delta CDE)$

Submit your answer as: and

4. Solving riders

ID is: 4154 Seed is: 2953

Algebraic triangles

Determine the size of $y$ , giving reasons for each of your statements.

INSTRUCTION: There is sometimes more than one way of solving a geometry problem. In this question, you must follow the structure given to you below. You should start by giving the answer for

\hat{R}

$\hat{R}$ in terms of

y

$y$ .

Answer:

\hat{R} =

$\hat{R} =$

∴

$\therefore$

y =

$y =$

°

$°$

expression

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

What do you know about triangles that have two equal sides? Once you've found $\hat{R}$ in terms of $y$ , write an equation for this triangle.

STEP: Find a missing angle in terms of

y

$y$
[−2 points ⇒ 3 / 5 points left]

The triangle has two equal sides, so we know that the angles opposite those sides will be equal.

\begin{matrix} ∴ \hat{R} = y (∠ s opp equal sides) \end{matrix}

$\begin{array}{c} \therefore \hat{R} = y \; \left(\angle \text{s opp equal sides}\right) \end{array}$

STEP: Write an equation for the triangle
[−2 points ⇒ 1 / 5 points left]

We know all the angles of the triangle in terms of $y$ , so we can write an equation using the sum of angles inside the triangle.

\begin{matrix} ∴ y + y + 3 y - 50 ° = 180 ° (sum of ∠ s in Δ) \end{matrix}

$\begin{array}{c} \therefore {\color{red}{y}}+{\color{red}{y}}+{\color{blue}{3y-50° }} =180° \;\left(\text{sum of } \angle \text{s in } \Delta\right) \end{array}$

STEP: Solve the equation to determine the value of

y

$y$
[−1 point ⇒ 0 / 5 points left]

\begin{aligned} ∴ 5 y - 50 ° & = 180 ° \\ ∴ 5 y - 50 ° + 50 ° & = 180 ° + 50 ° \\ ∴ 5 y & = 230 ° \\ ∴ \frac{5 y}{5} & = \frac{230 °}{5} \\ ∴ y & = 46 ° \end{aligned}

$\begin{aligned} \therefore 5y-50° & =180° \\ \therefore 5y-50° \color{brown}{+50°} & =180° \color{brown}{+50 °} \\ \therefore 5y&=230° \\ \therefore \color{teal}{\dfrac{\color{black}{5y}}{5}}&= \color{teal}{\dfrac{\color{black}{230°}}{5}}\\ \therefore y&=46° \end{aligned}$

Submit your answer as: andandand

ID is: 4154 Seed is: 1322

Algebraic triangles

Determine the size of $y$ , giving reasons for each of your statements.

INSTRUCTION: There is sometimes more than one way of solving a geometry problem. In this question, you must follow the structure given to you below. You should start by giving the answer for

\hat{C}

$\hat{C}$ in terms of

y

$y$ .

Answer:

\hat{C} =

$\hat{C} =$

∴

$\therefore$

y =

$y =$

°

$°$

expression

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

What do you know about triangles that have two equal sides? Once you've found $\hat{C}$ in terms of $y$ , write an equation for this triangle.

STEP: Find a missing angle in terms of

y

$y$
[−2 points ⇒ 3 / 5 points left]

The triangle has two equal sides, so we know that the angles opposite those sides will be equal.

\begin{matrix} ∴ \hat{C} = 2 y (∠ s opp equal sides) \end{matrix}

$\begin{array}{c} \therefore \hat{C} = 2y \; \left(\angle \text{s opp equal sides}\right) \end{array}$

STEP: Write an equation for the triangle
[−2 points ⇒ 1 / 5 points left]

We know all the angles of the triangle in terms of $y$ , so we can write an equation using the sum of angles inside the triangle.

\begin{matrix} ∴ 2 y + 2 y + 2 y + 12 ° = 180 ° (sum of ∠ s in Δ) \end{matrix}

$\begin{array}{c} \therefore {\color{red}{2y}}+{\color{red}{2y}}+{\color{blue}{2y+12° }} =180° \;\left(\text{sum of } \angle \text{s in } \Delta\right) \end{array}$

STEP: Solve the equation to determine the value of

y

$y$
[−1 point ⇒ 0 / 5 points left]

\begin{aligned} ∴ 6 y + 12 ° & = 180 ° \\ ∴ 6 y + 12 ° - 12 ° & = 180 ° - 12 ° \\ ∴ 6 y & = 168 ° \\ ∴ \frac{6 y}{6} & = \frac{168 °}{6} \\ ∴ y & = 28 ° \end{aligned}

$\begin{aligned} \therefore 6y+12° & =180° \\ \therefore 6y+12° \color{brown}{-12°} & =180° \color{brown}{-12 °} \\ \therefore 6y&=168° \\ \therefore \color{teal}{\dfrac{\color{black}{6y}}{6}}&= \color{teal}{\dfrac{\color{black}{168°}}{6}}\\ \therefore y&=28° \end{aligned}$

Submit your answer as: andandand

ID is: 4154 Seed is: 8298

Algebraic triangles

Determine the size of $z$ , giving reasons for each of your statements.

INSTRUCTION: There is sometimes more than one way of solving a geometry problem. In this question, you must follow the structure given to you below. You should start by giving the answer for

\hat{C}

$\hat{C}$ in terms of

z

$z$ .

Answer:

\hat{C} =

$\hat{C} =$

∴

$\therefore$

z =

$z =$

°

$°$

expression

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

What do you know about triangles that have two equal sides? Once you've found $\hat{C}$ in terms of $z$ , write an equation for this triangle.

STEP: Find a missing angle in terms of

z

$z$
[−2 points ⇒ 3 / 5 points left]

The triangle has two equal sides, so we know that the angles opposite those sides will be equal.

\begin{matrix} ∴ \hat{C} = 3 z (∠ s opp equal sides) \end{matrix}

$\begin{array}{c} \therefore \hat{C} = 3z \; \left(\angle \text{s opp equal sides}\right) \end{array}$

STEP: Write an equation for the triangle
[−2 points ⇒ 1 / 5 points left]

We know all the angles of the triangle in terms of $z$ , so we can write an equation using the sum of angles inside the triangle.

\begin{matrix} ∴ 3 z + 3 z + 5 z - 18 ° = 180 ° (sum of ∠ s in Δ) \end{matrix}

$\begin{array}{c} \therefore {\color{red}{3z}}+{\color{red}{3z}}+{\color{blue}{5z-18° }} =180° \;\left(\text{sum of } \angle \text{s in } \Delta\right) \end{array}$

STEP: Solve the equation to determine the value of

z

$z$
[−1 point ⇒ 0 / 5 points left]

\begin{aligned} ∴ 11 z - 18 ° & = 180 ° \\ ∴ 11 z - 18 ° + 18 ° & = 180 ° + 18 ° \\ ∴ 11 z & = 198 ° \\ ∴ \frac{11 z}{11} & = \frac{198 °}{11} \\ ∴ z & = 18 ° \end{aligned}

$\begin{aligned} \therefore 11z-18° & =180° \\ \therefore 11z-18° \color{brown}{+18°} & =180° \color{brown}{+18 °} \\ \therefore 11z&=198° \\ \therefore \color{teal}{\dfrac{\color{black}{11z}}{11}}&= \color{teal}{\dfrac{\color{black}{198°}}{11}}\\ \therefore z&=18° \end{aligned}$

Submit your answer as: andandand

Line
one	In $\Delta LMK$ and $\Delta KJL$ :
two	1. $M\hat{L}K = J\hat{K}L = 52°$ $\left(\text{given}\right)$
three	2. $LK \text{ is common}$
four	3. $\hat{J} = 78°$ $\left(\text{sum of } \angle \text{s in } \Delta\right)$
five	$\quad \therefore \hat{J} = \hat{M}$
six	$\therefore \Delta LMK \equiv \Delta KJL$
seven	$\therefore LM = KJ$ $\left(\Delta LMK \equiv \Delta KJL\right)$

Line
one	In $\Delta LMK$ and $\Delta KJL$ :
two	1. $M\hat{L}K = J\hat{K}L = 52°$ $\left(\text{given}\right)$
three	2. $LK \text{ is common}$
four	3. $\hat{J} = 78°$ $\left(\text{sum of } \angle \text{s in } \Delta\right)$
five	$\quad \therefore \hat{J} = \hat{M}$
six	$\therefore \Delta LMK \equiv \Delta KJL$ $\left(\text{SAA}\right)$
seven	$\therefore LM = KJ$ $\left(\Delta LMK \equiv \Delta KJL\right)$

Line
one	In $\Delta PRS$ and $\Delta SQP$ :
two	1. $\hat{Q} = 75°$ $\left(\text{sum of } \angle \text{s in } \Delta\right)$
three	$\quad \therefore \hat{Q} = \hat{R}$
four	2. $PSR = SPQ$ $\left(\text{sum of } \angle \text{s in } \Delta \right)$
five	3. $R\hat{P}S = Q\hat{S}P = 59°$ $\left(\text{given}\right)$
six	$\therefore \Delta PRS \equiv \Delta SQP$ $\left(\text{SAA}\right)$
seven	$\therefore PR = SQ$ $\left(\Delta PRS \equiv \Delta SQP\right)$

Line
one	In $\Delta PRS$ and $\Delta SQP$ :
two	1. $\hat{Q} = 75°$ $\left(\text{sum of } \angle \text{s in } \Delta\right)$
three	$\quad \therefore \hat{Q} = \hat{R}$
four	2. $PS \text{ is common}$
five	3. $R\hat{P}S = Q\hat{S}P = 59°$ $\left(\text{given}\right)$
six	$\therefore \Delta PRS \equiv \Delta SQP$ $\left(\text{SAA}\right)$
seven	$\therefore PR = SQ$ $\left(\Delta PRS \equiv \Delta SQP\right)$

Line
one	In $\Delta RSQ$ and $\Delta PSQ$ :
two	1. $QP^{2} = 12^{2} + 5^{2}$ $\left(\text{Pythagoras}\right)$
three	$\quad \therefore QP = 13$
four	$\quad \therefore QP = RQ$
five	2. $QS \text{ is common}$
six	3. $R\hat{S}Q = P\hat{S}Q = 90°$ $\left(\text{given}\right)$
seven	$\therefore \Delta RSQ \equiv \Delta PSQ$
eight	$\therefore R\hat{Q}S = P\hat{Q}S$ $\left(\Delta RSQ \equiv \Delta PSQ\right)$

A ✘	3. $AB \parallel DC$ $\;\;$ (given) $\therefore \Delta ABC \equiv \Delta DCB$ $\;\;$ (SAS)
$AB$ is parallel to $DC$ . But, this does not necessarily tell us anything about the length of $AB$ or $DC$ , becuase parallel lines do not have to be equal in length. This option does not prove that another pair of sides are equal.
B ✔	3. $\hat{D} = x$ $\;\;$ (given) $\;\;$ Also, $\hat{A} = A\hat{C}D = x$ $\;\;$ (alt $\angle$ s; $AB \parallel DC$ ) $\;\;$ $\therefore \hat{A} = \hat{D}$ $\;\;$ (both equal to $x$ ) $\therefore \Delta ABC \equiv \Delta DCB$ $\;\;$ (SAA)
This option uses geometry reasons correctly to prove that $\hat{A} = \hat{D}$ . This is a pair of matching angles in both triangles. So this proves that $\Delta ABC \equiv \Delta DCB$ , using SAA.
C ✘	3. $A\hat{B}D = x$ $\;\;$ (alt $\angle$ s; $AB \parallel DC$ ) $\;\;$ $\therefore A\hat{B}D = D\hat{C}A$ $\;\;$ (both equal to $x$ ) $\therefore \Delta ABC \equiv \Delta DCB$ $\;\;$ (SAA)
This option correctly shows that $A\hat{B}D$ is equal to $D\hat{C}A$ . But, these are not angles in the triangles that we are trying to prove are congruent.
D ✘	3. $A\hat{E}B = D\hat{E}C$ $\;\;$ (vert opp $\angle$ s) $\therefore \Delta ABC \equiv \Delta DCB$ $\;\;$ (SAA)
$A\hat{E}B$ is equal to $D\hat{E}C$ , because the angles are vertically opposite. But, these are not angles in the triangles that we are trying to prove are congruent.

A ✘	3. $TQ = TR$ $\;\;$ (given) $\therefore \Delta PQR \equiv \Delta SRQ$ $\;\;$ ( $90°$ HS)
$TQ$ is equal to $TR$ . But, these are not sides in the triangles that we are trying to prove are congruent. (They are only parts of the sides in those triangles, which is not enough.)
B ✘	3. $PR$ is the hypotenuse of $\Delta PQR$ $\;\;$ Also, $SQ$ is the hypotenuse of $\Delta SRQ$ . $\;\;$ $\therefore PR = SQ$ $\;\;$ (hypotenuse with common base) $\therefore \Delta PQR \equiv \Delta SRQ$ $\;\;$ ( $90°$ HS)
$PR$ and $SQ$ are hypotenuses of $\Delta PQR$ and $\Delta SRQ$ respectively. But, this does not mean that they have to be equal in length. Hypotenuse with common base does not refer to any of our geometry theorems. It is just a made up reason. So, this option does not prove that $PR = SQ$ .
C ✘	3. $P\hat{T}Q = S\hat{T}R$ $\;\;$ (vert opp $\angle$ s) $\therefore \Delta PQR \equiv \Delta SRQ$ $\;\;$ (SAA)
$P\hat{T}Q$ is equal to $S\hat{T}R$ , because the angles are vertically opposite. But, these are not angles in the triangles that we are trying to prove are congruent.
D ✔	3. $T\hat{Q}R = T\hat{R}Q$ $\;\;$ ( $\angle$ s opp equal sides) $\therefore \Delta PQR \equiv \Delta SRQ$ $\;\;$ (SAA)
This option uses geometry reasons correctly to prove that $T\hat{Q}R = T\hat{R}Q$ . This is a pair of matching angles in both triangles. So this proves that $\Delta PQR \equiv \Delta SRQ$ , using SAA.

Formal geometry proofs

1. Proofs of triangle and polygon theorems

Interior and exterior angles of a triangle

Interior and exterior angles of a triangle

Interior and exterior angles of a triangle

Interior angles of polygons

Interior angles of polygons

Interior angles of polygons

Exterior angles of polygons

Exterior angles of polygons

Exterior angles of polygons

Problems with polygons

Problems with polygons

Problems with polygons

2. Congruency in triangles

Prove congruency with a common side

Prove congruency with a common side

Prove congruency with a common side

Congruent triangles

Congruent triangles

Congruent triangles

Deductions from congruency: evaluating proofs

Deductions from congruency: evaluating proofs

Deductions from congruency: evaluating proofs

Prove simple congruency

Prove simple congruency

Prove simple congruency

Naming congruent triangles

Naming congruent triangles

Naming congruent triangles

Prove congruency with calculations

Prove congruency with calculations

Prove congruency with calculations

Identifying congruency in triangles

Identifying congruency in triangles

Identifying congruency in triangles

Congruency in circles

Congruency in circles

Congruency in circles

Consequences of congruency

Consequences of congruency

Consequences of congruency

Deductions from congruency

Deductions from congruency

Deductions from congruency

Consequences of congruency with calculations

Consequences of congruency with calculations

Consequences of congruency with calculations

Identify congruency in overlapping triangles

Identify congruency in overlapping triangles

Identify congruency in overlapping triangles

Deductions from congruency: overlapping triangles

Deductions from congruency: overlapping triangles

Deductions from congruency: overlapping triangles

Deductions from congruency: evaluating proofs

Deductions from congruency: evaluating proofs

Deductions from congruency: evaluating proofs

Deductions from congruency: overlapping triangles

Deductions from congruency: overlapping triangles

Deductions from congruency: overlapping triangles

3. Properties of quadrilaterals

Apply diagonal properties of a rhombus

Apply diagonal properties of a rhombus

Apply diagonal properties of a rhombus

Reasoning about properties of quadrilaterals

Reasoning about properties of quadrilaterals

Reasoning about properties of quadrilaterals

Identify all diagonal properties

Identify all diagonal properties

Identify all diagonal properties

Congruent triangles

Congruent triangles

Congruent triangles

Calculate angles between parallel lines

Calculate angles between parallel lines

Calculate angles between parallel lines

Proof: is the quadrilateral a parallelogram?

Proof: is the quadrilateral a parallelogram?

Proof: is the quadrilateral a parallelogram?

Choose one correct diagonal property